Denote $AB=c$, $BC=a$, $CA=b$, $p=\frac{a+b+c}{2}$, $\angle ABC=\beta$, $\angle BCA=\gamma$, $\angle CAB=\alpha$. Let $H_{1}$ be the foot of the perpendicular from $A'$ to line $B'C'$. Let $H_{2}$ be the foot of the perpendicular from $B'$ to line $A'C'$. Let $H_{3}$ be the foot of the perpendicular from $C'$ to line $A'B'$. Let $H_{4}$ be the foot of the perpendicular from $A'_{a}$ to line $B'_{a}C'_{a}$. Let $H_{5}$ be the foot of the perpendicular from $B'_{a}$ to line $A'_{a}C'_{a}$. Let $H_{6}$ be the foot of the perpendicular from $C'_{a}$ to line $A'_{a}B'_{a}$. Let $H_{7}$ be the foot of the perpendicular from $B'_{b}$ to line $A'_{b}C'_{b}$. Let $H_{8}$ be the foot of the perpendicular from $A'_{b}$ to line $B'_{b}C'_{b}$. Let $H_{9}$ be the foot of the perpendicular from $C'_{b}$ to line $A'_{b}B'_{b}$. Let $H_{10}$ be the foot of the perpendicular from $C'_{c}$ to line $A'_{c}B'_{c}$. Let $H_{11}$ be the foot of the perpendicular from $B'_{c}$ to line $A'_{c}C'_{c}$. Let $H_{12}$ be the foot of the perpendicular from $A'_{c}$ to line $C'_{c}B'_{c}$. It's well-known that: $$AB'=AC'=BC'_{c}=BA'_{c}=CB'_{b}=CA'_{b}=p-a$$$$BC'=BA'=CA'=CB'_{a}=AC'_{c}=AB'_{c}=p-b$$$$CB'=CA'=AB'_{b}=AC'_{b}=BA'_{a}=BC'_{a}=p-c$$$$CB'_{c}=CA'_{c}=AC'_{a}=AB'_{a}=BC'_{b}=BA'_{b}=p$$Thus we can express the following angles: $$\angle AB'C=\angle AC'B'=\angle B'A'C'=90^{\circ}-\frac{\alpha}{2}$$$$\angle BA'C'=\angle BC'A'=\angle A'B'C'=90^{\circ}-\frac{\beta}{2}$$$$\angle CB'A'=\angle CA'B'=\angle A'C'B'=90^{\circ}-\frac{\gamma}{2}$$$$\angle AB'_{b}C'_{b}=\angle AC'_{b}B'_{b}=\angle AB'_{c}C'_{c}=\angle AC'_{c}B'_{c}=\angle B'_{b}A'_{b}C'_{b}=\angle C'_{c}A'_{c}B'_{c}=\frac{\alpha}{2}$$$$\angle BC'_{c}A'_{c}=\angle BA'_{c}C'_{c}=\angle BA'_{a}C'_{a}=\angle BC'_{a}A'_{a}=\angle C'_{c}B'_{c}A'_{c}=\angle A'_{a}B'_{a}C'_{a}=\frac{\beta}{2}$$$$\angle CA'_{a}B'_{a}=\angle CB'_{a}A'_{a}=\angle CB'_{b}A'_{b}=\angle CA'_{b}B'_{b}=\angle A'_{a}C'_{a}B'_{a}=\angle B'_{b}C_{b}A'_{b}=\frac{\gamma}{2}$$Now we can calculate the altitudes in the four triangles by using the trigonometric functions and the Sine law as such: $$A'H_{1}=A'C'\sin\angle H_{1}C'A'=\sin\left(90^{\circ}-\frac{\gamma}{2}\right)\left(\frac{\sin\angle A'BC'}{\sin\angle BA'C'}BC'\right)=$$$$=\left(\cos\frac{\gamma}{2}\right)\frac{\sin\beta}{\sin(90^{\circ}-\frac{\beta}{2})}(p-b)=2\cos\frac{\gamma}{2}\sin\frac{\beta}{2}(p-b)$$We used that $\triangle A'H_{1}C'$ is a right triangle, the Sine law for $\triangle A'BC'$ and the double angle formula, i.e. $\sin 2x=2\sin x\cos x$. We will do a similar thing for the other altitudes, however we won't explain it in such detail as for the first one because this computation of $AH_{1}$ showcases the procedure. $$B'H_{2}=B'A'\sin(90^{\circ}-\frac{\alpha}{2})=\cos\frac{\alpha}{2}\times\frac{\sin\gamma}{\sin(90^{\circ}-\frac{\gamma}{2})}CB'=2\cos\frac{\alpha}{2}\sin\frac{\gamma}{2}(p-c)$$$$C'H_{3}=\sin(90^{\circ}-\frac{\beta}{2})B'C'=\cos\frac{\beta}{2}\times\frac{\sin\alpha}{\sin(90^{\circ}-\frac{\alpha}{2})}AC'=2\cos\frac{\gamma}{2}\sin\frac{\alpha}{2}(p-a)$$$$A'_{a}H_{4}=\sin\frac{\beta}{2} A'_{a}B'_{a}=\sin\frac{\beta}{2}\times \frac{\sin\gamma}{\sin\frac{\gamma}{2}}CA'_{a}=2\sin\frac{\beta}{2}\cos\frac{\gamma}{2}(p-b)$$$$B'_{a}H_{5}=\sin\frac{\gamma}{2}C'_{a}B'_{a}=\sin\frac{\gamma}{2}\times\frac{\sin{\alpha}}{\sin(90^{\circ}-\frac{\alpha}{2})}AB_{a}'=2\sin\frac{\alpha}{2}\sin\frac{\gamma}{2}p$$$$C'_{a}H_{6}=\sin\frac{\beta}{2}B'_{a}C'_{a}=\sin\frac{\beta}{2}\times\frac{\sin{\alpha}}{\sin(90^{\circ}-\frac{\alpha}{2})}AB'_{a}=2\sin\frac{\beta}{2}\sin\frac{\alpha}{2}p$$$$B'_{b}H_{7}=\sin\frac{\gamma}{2}C'_{b}B'_{b}=\sin\frac{\gamma}{2}\times \frac{\sin\alpha}{\sin\frac{\alpha}{2}}AB'_{b}=2\sin\frac{\gamma}{2}\cos\frac{\alpha}{2}(p-c)$$$$A'_{b}H_{8}=\sin\frac{\gamma}{2}A'_{b}C'_{b}=\sin\frac{\gamma}{2}\times \frac{\sin\beta}{\sin(90^{\circ}-\frac{\beta}{2})}BA'_{b}=2\sin\frac{\gamma}{2}\sin\frac{\beta}{2}p$$$$C'_{b}H_{9}=\sin\frac{\alpha}{2}C'_{b}A'_{b}=\sin\frac{\alpha}{2}\times \frac{\sin\beta}{\sin(90^{\circ}-\frac{\beta}{2})}BA'_{b}=2\sin\frac{\alpha}{2}\sin\frac{\beta}{2}p$$$$C'_{c}H_{10}=\sin\frac{\alpha}{2}C'_{c}A'_{c}=\sin\frac{\alpha}{2}\times\frac{\sin\beta}{\sin\frac{\beta}{2}}BC'_{c}=2\sin\frac{\alpha}{2}\cos\frac{\gamma}{2}(p-a)$$$$B'_{c}H_{11}=\sin\frac{\alpha}{2}B'_{c}A'_{c}=\sin\frac{\alpha}{2}\times\frac{\sin\gamma}{\sin(90^{\circ}-\frac{\gamma}{2})}CB'_{c}=2\sin\frac{\alpha}{2}\sin\frac{\gamma}{2}p$$$$A'_{c}H_{12}=\sin\frac{\beta}{2}B'_{c}A'_{c}=\sin\frac{\alpha}{2}\times\frac{\sin\gamma}{\sin(90^{\circ}-\frac{\gamma}{2})}CB'_{c}=2\sin\frac{\beta}{2}\sin\frac{\gamma}{2}p$$$$\Longrightarrow A'H_{1}=A'_{a}H_{4},\quad B'H_{2}=B'_{b}H_{7},\quad C'H_{3}=C'_{c}H_{10}$$$$\Longrightarrow B'_{a}H_{5}=B'_{c}H_{11},\quad C'_{a}H_{6}=C'_{b}H_{9},\quad A'_{b}H_{8}=A'_{c}H_{12}$$Therefore we can (maybe, we haven't shown yet) have a maximum of $6$ different lengths for the altitudes. However, by using the double angle formulas $\sin^2 x=\frac{1-\cos 2x}{2}$, $\cos^2 x=\frac{1+\cos 2x}{2}$ and the Cosine law for $\triangle ABC$ to express $\cos\alpha=\frac{-a^2+b^2+c^2}{2bc}$ (and the cyclic expressions for $\beta$ and $\gamma$ respectively) we can express the lengths of the altitudes in another way. Note that if we have two altitudes with lengths $h_{1}$ and $h_{2}$ we know that $h_{1}=h_{2}\iff h_{1}^2=h_{2}^2$ because $h_{1}$ and $h_{2}$ are non-negative. Therefore we can just work with the squares of the lengths of the altitudes. Denote $A'H_{1}=h_{1}$, $B'H_{2}=h_{2}$, $C'H_{3}=h_{3}$, $B'_{a}H_{5}=h_{4}$, $C'_{a}H_{6}=h_{5}$, $A'_{b}H_{8}=h_{6}$ for shortness. Then: $$h_{1}^2=4\cos^2\frac{\gamma}{2}\sin^2\frac{\beta}{2}(p-b)^2=4\left(\frac{1+\cos\gamma}{2}\right)\left(\frac{1-\cos\beta}{2}\right)\left(\frac{a-b+c}{2}\right)^2=$$$$=\left(1+\frac{a^2+b^2-c^2}{2ab}\right)\left(1-\frac{a^2-b^2+c^2}{2ac}\right)\left(\frac{a-b+c}{2}\right)^2=$$$$=\frac{1}{16a^2bc}\left((a+b)^2-c^2\right)\left(b^2-(a-c)^2\right)\left(a-b+c\right)^2=$$$$=\frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{16abc}\times \frac{(a+b-c)(a-b+c)}{a}$$Analogically: $$h_{2}^2=\frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{16abc}\times \frac{(a+b-c)(-a+b+c)}{b}$$$$h_{3}^2=\frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{16abc}\times \frac{(-a+b+c)(a-b+c)}{c}$$Now, for $h_{4},h_{5},h_{6}$ we have that: $$h_{4}^2=4\sin^2\frac{\alpha}{2}\sin^2\frac{\gamma}{2}p^2=4\left(\frac{1-\cos\alpha}{2}\right)\left(\frac{1-\cos\gamma}{2}\right)\left(\frac{a+b+c}{2}\right)^2=$$$$=\left(1-\frac{-a^2+b^2+c^2}{2bc}\right)\left(1-\frac{a^2+b^2-c^2}{2ab}\right)\left(\frac{a+b+c}{2}\right)^2=$$$$=\frac{1}{16ab^2c}(a^2-(b-c)^2)(c^2-(a-b)^2)(a+b+c)^2=$$$$=\frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{16abc}\times \frac{(a+b+c)(a-b+c)}{b}$$Analogically: $$h_{5}^2=\frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{16abc}\times \frac{(a+b+c)(a+b-c)}{c}$$$$h_{6}^2=\frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{16abc}\times \frac{(a+b+c)(-a+b+c)}{a}$$Notice that the expression $$T=\frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{16abc}$$appears in all altitudes and is non-zero as $a,b,c,$ are the lengths of the sides of $\triangle ABC$, so $a+b>c$, $a+c>b$, $b+c>a\Longrightarrow T\neq 0$. Therefore we may rewrite the expressions as follows: $$h_{1}^2=T\times \frac{(a+b-c)(a-b+c)}{a}$$$$h_{2}^2=T\times \frac{(a+b-c)(-a+b+c)}{b}$$$$h_{3}^2=T\times \frac{(-a+b+c)(a-b+c)}{c}$$$$h_{4}^2=T\times \frac{(a+b+c)(a-b+c)}{b}$$$$h_{5}^2=T\times \frac{(a+b+c)(a+b-c)}{c}$$$$h_{6}^2=T\times \frac{(a+b+c)(-a+b+c)}{a}$$Now we're ready to solve the two parts of the problem: $\textbf{a)}$ We showed that there are six pairs of equal lengths of altitudes, so the maximal number of different lengths in at most $6$. On the other hand, if $a=2,b=3,c=4$ we have that: $$h_{1}^2=T\times \frac{3}{2}$$$$h_{2}^2=T\times \frac{5}{3}$$$$h_{3}^2=T\times \frac{15}{4}$$$$h_{4}^2=T\times 9$$$$h_{5}^2=T\times \frac{9}{4}$$$$h_{6}^2=T\times \frac{45}{2}$$We mentioned that $T\neq 0$ and that two altitudes have equal lengths iff their squares are equal, but here we've shown that they're all different, so the maximal number of different lengths of the altitudes is $6$ and is achieved for example when $a=2,b=3,c=4$.\newline\newline $\textbf{Answer: \boxed{6}}$ $\textbf{b)}$ We showed an example where the number of different lengths is $6$. We'll show examples for there being $5,4,3,2$ different lengths and show that we can't have all lengths of the altitudes be equal. $\textbf{5 different lengths}$ If $a=3,b=4,c=5$ we have: $$h_{1}^2=T\times \frac{8}{3}$$$$h_{2}^2=T\times 3$$$$h_{3}^2=T\times \frac{24}{5}$$$$h_{4}^2=T\times 12$$$$h_{5}^2=T\times \frac{24}{5}$$$$h_{6}^2=T\times 24$$Here $h_{3}^2=h_{5}^2\Longrightarrow h_{3}=h_{5}$ and so we get $5$ different lengths because $T\neq 0$ (I will no longer mention this, as it's getting repetitive). In general, if we have $ABC$ be a non-isosceles right triangle we will have $5$ different lengths. $\textbf{4 different lengths}$ If $a=2, b=2, c=3$ we have: $$h_{1}^2=T\times \frac{3}{2}$$$$h_{2}^2=T\times \frac{3}{2}$$$$h_{3}^2=T\times 3$$$$h_{4}^2=T\times \frac{21}{2}$$$$h_{5}^2=T\times \frac{7}{3}$$$$h_{6}^2=T\times \frac{21}{2}$$Here we have $h_{1}=h_{2}$ and $h_{4}=h_{6}$ and we get $4$ different lengths. In general, if $\triangle ABC$ is isosceles, but not equilateral, then we'd have $4$ different lengths. $\textbf{3 different lengths}$ If $a=1, b=1, c=\sqrt{2}$ we have: $$h_{1}^2=T\times (2-\sqrt{2})\sqrt{2}$$$$h_{2}^2=T\times (2-\sqrt{2})\sqrt{2}$$$$h_{3}^2=T\times \sqrt{2}$$$$h_{4}^2=T\times (2+\sqrt{2})\sqrt{2}$$$$h_{5}^2=T\times \sqrt{2}$$$$h_{6}^2=T\times (2+\sqrt{2})\sqrt{2}$$Here $h_{1}=h_{2}$, $h_{4}=h_{6}$ and $h_{3}=h_{5}$ and we get $3$ different lengths In general, if $\triangle ABC$ is isosceles and a right triangle, then we get $3$ different lengths. $\textbf{2 different lengths}$ If $a=b=c=1$ we have: $$h_{1}^2=T$$$$h_{2}^2=T$$$$h_{3}^2=T$$$$h_{4}^2=T\times 3$$$$h_{5}^2=T\times 3$$$$h_{6}^2=T\times 3$$Here $h_{1}=h_{2}=h_{3}\neq h_{4}=h_{5}=h_{6}$ and we get 2 different lengths. In general, if $\triangle ABC$ is equilateral, we'd get $2$ different lengths. $\textbf{only 1 length}$ Assume that a $\triangle ABC$ exists, such that $a,b,c>0$ and all of the altitudes are equal. Then we have that: $$h_{1}^2=h_{6}^2\Longrightarrow T\times \frac{(a+b-c)(a-b+c)}{a}=\frac{(a+b+c)(-a+b+c)}{a}$$$$\Longrightarrow (a-b+c)(a+b-c)=(a+b+c)(-a+b+c)\Longrightarrow a^2=b^2+c^2$$$$h_{2}^2=h_{4}^2\Longrightarrow T\times \frac{(a+b-c)(-a+b+c)}{b}=\frac{(a+b+c)(a-b+c)}{b}$$$$\Longrightarrow (a+b-c)(-a+b-c)=(a+b+c)(a-b+c)\Longrightarrow b^2=a^2+c^2$$$$\Longrightarrow a^2=b^2+c^2=(a^2+c^2)+c^2\Longrightarrow 2c^2=0\Longrightarrow c=0$$This, however, is clearly impossible, so we reach a contradiction. Thus all possible numbers of different lengths are $2,3,4,5,6$. $\textbf{Answer: \boxed{2,3,4,5,6}}$

The possible numbers of different lengths are 2, 3, 4, 5, and 6. Let $I,I_A,I_B,I_C$ be the incenter and three excenters of $ABC$. Claim. $\triangle BIC \sim \triangle B_a'A_a'C_a'$, and symmetric variants. Proof. We have \[ \angle A_a'C_a'B_a' = \angle A_a'I_AI_B = 90^\circ- \angle I_ACB = \angle BCI, \]and similarly $\angle C_a'B_a'A_a' = \angle IBC$. This establishes $\triangle BIC \sim \triangle B_a'A_a'C_a'$, and symmetric angle chasing can be done for the other two similarities. $\blacksquare$ Let $\rho_A$ be the ratio of similarity of $\triangle B_a'A_a'C_a'$ to $\triangle BIC$; define $\rho_B,\rho_C$ similarly. Let $\alpha,\beta,\gamma$ be half the angles of $\angle BAC, \angle CBA, \angle ACB$ respectively. Claim. $\rho_A = 2\cos\beta\cos\gamma$, and symmetric variants. Proof. The circumradius of $\triangle BIC$ is equal to the distance between the midpoint of arc $\widehat{BC}$ in $(ABC)$ and $B$, which is $2R\sin\alpha$, where $R$ is the circumradius of $(ABC)$. The circumradius of $\triangle B_a'A_a'C_a'$ is the $A$-exradius, which is $4R\sin\alpha\cos\beta\cos\gamma$. Dividing the two circumradii gives the claim. $\blacksquare$ Let $d_a,e_a,f_a$ be the lengths of the altitudes, respectively, from $A_a',B_a',C_a'$ in $\triangle A_a'B_a'C_a'$, and define $d_b,e_b,f_b$ and $d_c,e_c,f_c$ similarly; let $d,e,f$ be the lengths of the altitudes from $A',B',C'$ in $\triangle A'B'C'$. Let $r$ be the inradius of $\triangle ABC$. Claim. $d_a=d$, $d_b=e$, and $d_c=f$. Proof. Since $d_a$ and $IA'$ are corresponding parts in $\triangle B_a'A_a'C_a'$ and $\triangle BIC$, we have \[ \frac{d_a}{r} = \rho_A \; \iff \; d_a = 2r\cos\beta \cos\gamma. \]Now, using the fact that the circumradius of a triangle is equal to the ratio of the product of its sides and 4 times its area, we obtain \begin{align*} r &= \frac{B'C'\cdot C'A' \cdot A'B'}{4S_{\triangle A'B'C'}} \\ &= \frac{B'C'\cdot C'A' \cdot A'B'}{4\left(\frac12 d \cdot B'C'\right)} \\ &= \frac{2BA'\sin\beta \cdot 2CA'\sin\gamma}{2d}, \end{align*}which rearranges to \[ d = \frac{2BA'\cdot CA' \sin\beta\sin\gamma}{r}. \]We'd like to show that this quantity is equal to $2r\cos\beta\cos\gamma$. But this desired equality is equivalent to \[ \tan\beta \tan\gamma = \frac{r}{BA'} \cdot \frac{r}{CA'}, \]which is true as $\tan\beta = r/BA'$ and $\tan\gamma = r/CA'$. Thus, we've shown that $d_a=d$, and the other two are analogous. $\blacksquare$ Let $a,b,c$ be the lengths of $BC,CA,AB$. We can compute $e_a$ in a similar manner. By similar triangles, \[ \frac{e_a}{\text{dist}(B, IC)} = \rho_A, \]and $\text{dist}(B, IC) = a\sin\gamma$. So, \[ e_a = 2a\cos\beta\sin\gamma\cos\gamma = a\cos\gamma\sin(2\beta). \]Upon performing completely analogous calculations for the other altitudes, we obtain similar expressions. (Note: due to the labeling of the problem, I might have screwed up the indices; this shouldn't impede the validity of the solution since we're only interested in the set of lengths.) \[ \begin{tabular}{ccc} $d_a = 2r\cos\beta\cos\gamma = d$ & $d_b = 2r\cos\gamma\cos\alpha = e$ & $d_c = 2r\cos\alpha\cos\beta = f$ \\ $e_a = a\cos\beta\sin(2\gamma)$ & $e_b = b\cos\gamma\sin(2\alpha)$ & $e_c = c\cos\alpha\sin(2\beta)$ \\ $f_a = a\cos\gamma\sin(2\beta)$ & $f_b = b\cos\alpha\sin(2\gamma)$ & $f_c = c\cos\beta\sin(2\alpha)$ \\ \end{tabular} \] Claim. $f_a=e_b$, $f_b=e_c$, and $f_c=e_a$. Proof. By the law of sines in $\triangle ABC$, we have \[ \frac{a}{\sin(2\alpha)} = \frac{b}{\sin(2\beta)} \; \iff \; a\sin(2\beta) = b\sin(2\alpha), \]so $f_a=e_b$. The other two equalities follow analogously. $\blacksquare$ Now, we compare the elements of the sets $\{d_a,d_b,d_c\}$ and $\{e_a,e_b,e_c\}$. Note that we can cancel out cosines of $\alpha$, etc. without qualification since the half-angles are strictly acute. Also, in what follows, the well-known triangle half-angle identities will be used repeatedly: \[ \sin^2\alpha = \frac{(s-b)(s-c)}{bc}, \; \cos^2\alpha = \frac{s(s-a)}{bc}, \; \tan^2\alpha = \frac{(s-b)(s-c)}{s(s-a)}. \] Claim. $e_a$ is never equal to $d_a$ and $d_c$; $e_b$ is never equal to $d_b$ and $d_a$; $e_c$ is never equal to $d_c$ and $d_b$. Proof. Comparing $e_a$ and $d_a$, we have \begin{align*} d_a = e_a \; &\iff \; 2r\cos\gamma = a\sin(2\gamma) = 2a\sin\gamma\cos\gamma \\ &\iff \; r = a\sin\gamma = a\cdot \frac{r}{IC} \\ &\iff \; a = IC \\ &\iff \; \angle BIC = \beta, \end{align*}which is impossible since $\angle BIC = 90^\circ+ \angle BAC/2$ is always obtuse while $\beta = \angle ABC/2$ is always acute. Comparing $e_a$ and $d_c$, we have \begin{align*} e_a = d_c \; &\iff \; a\sin(2\gamma) = 2r\cos\alpha \\ &\iff \; a\sin\gamma\cos\gamma = r\cos\alpha \\ &\iff \; a^2 \cdot \frac{(s-a)(s-b)}{ab}\cdot\frac{s(s-c)}{ab} = r^2\cdot\frac{s(s-a)}{bc} \\ &\iff \; \frac{(s-b)(s-c)}{b} = \frac{r^2}{c} \\ &\iff \; \frac{c}{b} = \frac{r^2}{(s-b)(s-c)} = \tan\beta\tan\gamma \\ &\iff \; \frac{c^2}{b^2} = \frac{(s-c)(s-a)}{s(s-b)} \cdot \frac{(s-a)(s-b)}{s(s-c)} = \frac{(s-a)^2}{s^2} \\ &\iff \; \frac{c}{b} = \frac{s-a}{s} = \frac{b+c-a}{a+b+c} \\ &\iff \; ac+bc+c^2 = b^2+bc-ab \\ &\iff \; a(b+c) = b^2-c^2 \\ &\iff \; a=b-c, \end{align*}which violates the triangle inequality. The comparisons for $e_b$ and $e_c$ are analogous. $\blacksquare$ Claim. $e_a=d_b$ if and only if $\angle ABC$ is right; $e_b=d_c$ if and only if $\angle BCA$ is right; $e_c=d_a$ if and only if $\angle CAB$ is right. Proof. Comparing $e_a$ and $d_b$, we have \begin{align*} e_a = d_b \; &\iff \; a\cos\beta\cdot2\sin\gamma\cos\gamma = 2r\cos\gamma\cos\alpha \\ &\iff \; a\cos\beta\sin\gamma = r\cos\alpha \\ &\iff \; a^2\cdot \frac{s(s-b)}{ac} \cdot \frac{(s-a)(s-b)}{ab} = r^2 \cdot \frac{s(s-a)}{bc} \\ &\iff \; (s-b)^2 = r^2 \\ &\iff \; s-b=r \\ &\iff \; \beta = 45^\circ, \end{align*}which means that $\angle ABC$ must be right. The other two assertions for $e_b$ and $e_c$ follow analogously. $\blacksquare$ Claim. $e_b=e_c$ if and only if $a=c$; $e_c=e_a$ if and only if $a=b$; and $e_a=e_b$ if and only if $b=c$. Proof. We have \begin{align*} e_b = e_c \; &\iff \; b\cos\gamma(2\sin\alpha\cos\alpha) = c\cos\alpha(2\sin\beta\cos\beta) \\ &\iff \; b\cos\gamma\sin\alpha = c\sin\beta\cos\beta \\ &\iff \; b^2 \cdot \frac{s(s-c)}{ab} \cdot \frac{(s-b)(s-c)}{bc} = c^2 \cdot \frac{(s-c)(s-a)}{ca} \cdot \frac{s(s-b)}{ac} \\ &\iff \; \frac{s-c}{c} = \frac{s-a}{a} \\ &\iff \; \frac sc = \frac sa \\ &\iff \; a = c, \end{align*}and the other two assertions are similar. $\blacksquare$ Claim. $d_b=d_c$ if and only if $b=c$, and similarly for the other pairs. Proof. Again, note that $\cos\alpha$ is nonzero since $\alpha$ is acute, so we have \[ d_b = d_c \; \iff \; \cos\gamma = \cos\beta. \]Since $\gamma, \beta$ are acute and $\cos(\bullet)$ is strictly decreasing on $(0, 90^\circ)$, the equation is equivalent to $\gamma=\beta$, or $b=c$. $\blacksquare$ We see that the only conditions governing the size of the set of altitude lengths is whether $\triangle ABC$ is right and whether it is isosceles. The following table summarizes our results, where the necessary and sufficient condition is placed in the cell corresponding to the two quantities in the row and column headings being equal ($\bullet$ represents ``cannot happen"): \[ \begin{tabular}{c||c|c|c|c|c|c} & $d_a$ & $d_b$ & $d_c$ & $e_a$ & $e_b$ & $e_c$ \\ \hline\hline $d_a$ & X & $a=b$ & $a=c$ & $\bullet$ & $\bullet$ & $\angle A = 90^\circ$ \\ \hline $d_b$ & & X & $b=c$ & $\angle B = 90^\circ$ & $\bullet$ & $\bullet$ \\ \hline $d_c$ & & & X & $\bullet$ & $\angle C = 90^\circ$ & $\bullet$ \\ \hline $e_a$ & & & & X & $b=c$ & $a=b$ \\ \hline $e_b$ & & & & & X & $a=c$ \\ \hline $e_c$ & & & & & & X \\ \end{tabular} \]We now do casework on the two relevant conditions. In what follows, ``triangle" refers to $\triangle ABC$. \begin{enumerate} \item Triangle is right; assume $\angle A = 90^\circ$. \begin{enumerate} \item Triangle is non-isosceles. Then $d_a=e_c$ while all the other altitudes are different, leading to 5 distinct lengths. \item Triangle is isosceles (then it must be that $b=c$, and the triangle is not equilateral). Then $d_a=e_c$, $d_b=d_c$, and $e_a=e_b$, leading to 3 distinct lengths. \end{enumerate} \item Triangle has no right angles. \begin{enumerate} \item Triangle is isosceles but not equilateral; assume $b=c\neq a$. Then $d_b=d_c$ and $e_a=e_b$, leading to 4 distinct lengths. \item Triangle is equilateral ($a=b=c$). Then $(d_a,d_b,d_c)$ are all equal and $(e_a,e_b,e_c)$ are all equal, leading to 2 distinct lengths. \item Triangle is non-isosceles. Then all 6 lengths are distinct. \end{enumerate} \end{enumerate} This exhausts all the cases, and in conclusion there can be 2, 3, 4, 5, and 6 different lengths.

The possible numbers of different lengths are 2, 3, 4, 5, and 6. Let $I,I_A,I_B,I_C$ be the incenter and three excenters of $ABC$. Claim. $\triangle BIC \sim \triangle B_a'A_a'C_a'$, and symmetric variants. Proof. We have \[ \angle A_a'C_a'B_a' = \angle A_a'I_AI_B = 90^\circ- \angle I_ACB = \angle BCI, \]and similarly $\angle C_a'B_a'A_a' = \angle IBC$. This establishes $\triangle BIC \sim \triangle B_a'A_a'C_a'$, and symmetric angle chasing can be done for the other two similarities. $\blacksquare$ Let $\rho_A$ be the ratio of similarity of $\triangle B_a'A_a'C_a'$ to $\triangle BIC$; define $\rho_B,\rho_C$ similarly. Let $\alpha,\beta,\gamma$ be half the angles of $\angle BAC, \angle CBA, \angle ACB$ respectively. Claim. $\rho_A = 2\cos\beta\cos\gamma$, and symmetric variants. Proof. The circumradius of $\triangle BIC$ is equal to the distance between the midpoint of arc $\widehat{BC}$ in $(ABC)$ and $B$, which is $2R\sin\alpha$, where $R$ is the circumradius of $(ABC)$. The circumradius of $\triangle B_a'A_a'C_a'$ is the $A$-exradius, which is $4R\sin\alpha\cos\beta\cos\gamma$. Dividing the two circumradii gives the claim. $\blacksquare$ Let $d_a,e_a,f_a$ be the lengths of the altitudes, respectively, from $A_a',B_a',C_a'$ in $\triangle A_a'B_a'C_a'$, and define $d_b,e_b,f_b$ and $d_c,e_c,f_c$ similarly; let $d,e,f$ be the lengths of the altitudes from $A',B',C'$ in $\triangle A'B'C'$. Let $r$ be the inradius of $\triangle ABC$. Claim. $d_a=d$, $d_b=e$, and $d_c=f$. Proof. Since $d_a$ and $IA'$ are corresponding parts in $\triangle B_a'A_a'C_a'$ and $\triangle BIC$, we have \[ \frac{d_a}{r} = \rho_A \; \iff \; d_a = 2r\cos\beta \cos\gamma. \]Now, using the fact that the circumradius of a triangle is equal to the ratio of the product of its sides and 4 times its area, we obtain \begin{align*} r &= \frac{B'C'\cdot C'A' \cdot A'B'}{4S_{\triangle A'B'C'}} \\ &= \frac{B'C'\cdot C'A' \cdot A'B'}{4\left(\frac12 d \cdot B'C'\right)} \\ &= \frac{2BA'\sin\beta \cdot 2CA'\sin\gamma}{2d}, \end{align*}which rearranges to \[ d = \frac{2BA'\cdot CA' \sin\beta\sin\gamma}{r}. \]We'd like to show that this quantity is equal to $2r\cos\beta\cos\gamma$. But this desired equality is equivalent to \[ \tan\beta \tan\gamma = \frac{r}{BA'} \cdot \frac{r}{CA'}, \]which is true as $\tan\beta = r/BA'$ and $\tan\gamma = r/CA'$. Thus, we've shown that $d_a=d$, and the other two are analogous. $\blacksquare$ Let $a,b,c$ be the lengths of $BC,CA,AB$. We can compute $e_a$ in a similar manner. By similar triangles, \[ \frac{e_a}{\text{dist}(B, IC)} = \rho_A, \]and $\text{dist}(B, IC) = a\sin\gamma$. So, \[ e_a = 2a\cos\beta\sin\gamma\cos\gamma = a\cos\gamma\sin(2\beta). \]Upon performing completely analogous calculations for the other altitudes, we obtain similar expressions. (Note: due to the labeling of the problem, I might have screwed up the indices; this shouldn't impede the validity of the solution since we're only interested in the set of lengths.) \[ \begin{tabular}{ccc} $d_a = 2r\cos\beta\cos\gamma = d$ & $d_b = 2r\cos\gamma\cos\alpha = e$ & $d_c = 2r\cos\alpha\cos\beta = f$ \\ $e_a = a\cos\beta\sin(2\gamma)$ & $e_b = b\cos\gamma\sin(2\alpha)$ & $e_c = c\cos\alpha\sin(2\beta)$ \\ $f_a = a\cos\gamma\sin(2\beta)$ & $f_b = b\cos\alpha\sin(2\gamma)$ & $f_c = c\cos\beta\sin(2\alpha)$ \\ \end{tabular} \] Claim. $f_a=e_b$, $f_b=e_c$, and $f_c=e_a$. Proof. By the law of sines in $\triangle ABC$, we have \[ \frac{a}{\sin(2\alpha)} = \frac{b}{\sin(2\beta)} \; \iff \; a\sin(2\beta) = b\sin(2\alpha), \]so $f_a=e_b$. The other two equalities follow analogously. $\blacksquare$ Now, we compare the elements of the sets $\{d_a,d_b,d_c\}$ and $\{e_a,e_b,e_c\}$. Note that we can cancel out cosines of $\alpha$, etc. without qualification since the half-angles are strictly acute. Also, in what follows, the well-known triangle half-angle identities will be used repeatedly: \[ \sin^2\alpha = \frac{(s-b)(s-c)}{bc}, \; \cos^2\alpha = \frac{s(s-a)}{bc}, \; \tan^2\alpha = \frac{(s-b)(s-c)}{s(s-a)}. \] Claim. $e_a$ is never equal to $d_a$ and $d_c$; $e_b$ is never equal to $d_b$ and $d_a$; $e_c$ is never equal to $d_c$ and $d_b$. Proof. Comparing $e_a$ and $d_a$, we have \begin{align*} d_a = e_a \; &\iff \; 2r\cos\gamma = a\sin(2\gamma) = 2a\sin\gamma\cos\gamma \\ &\iff \; r = a\sin\gamma = a\cdot \frac{r}{IC} \\ &\iff \; a = IC \\ &\iff \; \angle BIC = \beta, \end{align*}which is impossible since $\angle BIC = 90^\circ+ \angle BAC/2$ is always obtuse while $\beta = \angle ABC/2$ is always acute. Comparing $e_a$ and $d_c$, we have \begin{align*} e_a = d_c \; &\iff \; a\sin(2\gamma) = 2r\cos\alpha \\ &\iff \; a\sin\gamma\cos\gamma = r\cos\alpha \\ &\iff \; a^2 \cdot \frac{(s-a)(s-b)}{ab}\cdot\frac{s(s-c)}{ab} = r^2\cdot\frac{s(s-a)}{bc} \\ &\iff \; \frac{(s-b)(s-c)}{b} = \frac{r^2}{c} \\ &\iff \; \frac{c}{b} = \frac{r^2}{(s-b)(s-c)} = \tan\beta\tan\gamma \\ &\iff \; \frac{c^2}{b^2} = \frac{(s-c)(s-a)}{s(s-b)} \cdot \frac{(s-a)(s-b)}{s(s-c)} = \frac{(s-a)^2}{s^2} \\ &\iff \; \frac{c}{b} = \frac{s-a}{s} = \frac{b+c-a}{a+b+c} \\ &\iff \; ac+bc+c^2 = b^2+bc-ab \\ &\iff \; a(b+c) = b^2-c^2 \\ &\iff \; a=b-c, \end{align*}which violates the triangle inequality. The comparisons for $e_b$ and $e_c$ are analogous. $\blacksquare$ Claim. $e_a=d_b$ if and only if $\angle ABC$ is right; $e_b=d_c$ if and only if $\angle BCA$ is right; $e_c=d_a$ if and only if $\angle CAB$ is right. Proof. Comparing $e_a$ and $d_b$, we have \begin{align*} e_a = d_b \; &\iff \; a\cos\beta\cdot2\sin\gamma\cos\gamma = 2r\cos\gamma\cos\alpha \\ &\iff \; a\cos\beta\sin\gamma = r\cos\alpha \\ &\iff \; a^2\cdot \frac{s(s-b)}{ac} \cdot \frac{(s-a)(s-b)}{ab} = r^2 \cdot \frac{s(s-a)}{bc} \\ &\iff \; (s-b)^2 = r^2 \\ &\iff \; s-b=r \\ &\iff \; \beta = 45^\circ, \end{align*}which means that $\angle ABC$ must be right. The other two assertions for $e_b$ and $e_c$ follow analogously. $\blacksquare$ Claim. $e_b=e_c$ if and only if $a=c$; $e_c=e_a$ if and only if $a=b$; and $e_a=e_b$ if and only if $b=c$. Proof. We have \begin{align*} e_b = e_c \; &\iff \; b\cos\gamma(2\sin\alpha\cos\alpha) = c\cos\alpha(2\sin\beta\cos\beta) \\ &\iff \; b\cos\gamma\sin\alpha = c\sin\beta\cos\beta \\ &\iff \; b^2 \cdot \frac{s(s-c)}{ab} \cdot \frac{(s-b)(s-c)}{bc} = c^2 \cdot \frac{(s-c)(s-a)}{ca} \cdot \frac{s(s-b)}{ac} \\ &\iff \; \frac{s-c}{c} = \frac{s-a}{a} \\ &\iff \; \frac sc = \frac sa \\ &\iff \; a = c, \end{align*}and the other two assertions are similar. $\blacksquare$ Claim. $d_b=d_c$ if and only if $b=c$, and similarly for the other pairs. Proof. Again, note that $\cos\alpha$ is nonzero since $\alpha$ is acute, so we have \[ d_b = d_c \; \iff \; \cos\gamma = \cos\beta. \]Since $\gamma, \beta$ are acute and $\cos(\bullet)$ is strictly decreasing on $(0, 90^\circ)$, the equation is equivalent to $\gamma=\beta$, or $b=c$. $\blacksquare$ We see that the only conditions governing the size of the set of altitude lengths is whether $\triangle ABC$ is right and whether it is isosceles. The following table summarizes our results, where the necessary and sufficient condition is placed in the cell corresponding to the two quantities in the row and column headings being equal ($\bullet$ represents ``cannot happen"): \[ \begin{tabular}{c||c|c|c|c|c|c} & $d_a$ & $d_b$ & $d_c$ & $e_a$ & $e_b$ & $e_c$ \\ \hline\hline $d_a$ & X & $a=b$ & $a=c$ & $\bullet$ & $\bullet$ & $\angle A = 90^\circ$ \\ \hline $d_b$ & & X & $b=c$ & $\angle B = 90^\circ$ & $\bullet$ & $\bullet$ \\ \hline $d_c$ & & & X & $\bullet$ & $\angle C = 90^\circ$ & $\bullet$ \\ \hline $e_a$ & & & & X & $b=c$ & $a=b$ \\ \hline $e_b$ & & & & & X & $a=c$ \\ \hline $e_c$ & & & & & & X \\ \end{tabular} \]We now do casework on the two relevant conditions. In what follows, ``triangle" refers to $\triangle ABC$. \begin{enumerate} \item Triangle is right; assume $\angle A = 90^\circ$. \begin{enumerate} \item Triangle is non-isosceles. Then $d_a=e_c$ while all the other altitudes are different, leading to 5 distinct lengths. \item Triangle is isosceles (then it must be that $b=c$, and the triangle is not equilateral). Then $d_a=e_c$, $d_b=d_c$, and $e_a=e_b$, leading to 3 distinct lengths. \end{enumerate} \item Triangle has no right angles. \begin{enumerate} \item Triangle is isosceles but not equilateral; assume $b=c\neq a$. Then $d_b=d_c$ and $e_a=e_b$, leading to 4 distinct lengths. \item Triangle is equilateral ($a=b=c$). Then $(d_a,d_b,d_c)$ are all equal and $(e_a,e_b,e_c)$ are all equal, leading to 2 distinct lengths. \item Triangle is non-isosceles. Then all 6 lengths are distinct. \end{enumerate} \end{enumerate} This exhausts all the cases, and in conclusion there can be 2, 3, 4, 5, and 6 different lengths.

I preferred working with just $a,b,c$ and not trigonometric functions as the final forms of the lengths of the altitudes, because it takes much more effort to look through the cases, whereas for the $a,b,c,$ representation we can just give examples which reach the desired result for $2$ to $6$ and be done. Any correct solution is still a good solution though.

if you stick with the trig method, like in ppanther's solution you need to consider a lot of cases, which are not trivial and for my approach, I think the idea to square the altitudes in order to represent them metrically is really clever and non-standard (at least I haven't seen it anywhere as a method)