It's easy to see that it must be +1 or -1. Start with any point o, take a regular tetrahedron abcd whose center is o and apply the condition to abcd, abco, abdo, acdo, bcdo. The centers of abco, abdo, acdo, bcdo form a tetrahedron whose center is also o. With all these equations we get f(o)=+1 or -1. Now suppose there is a point o with -1. Again take a regular tetrahedron abcd whose center is o. Suppose f(a)=-1 and f(b)=f(c)=f(d)=1, the other case is analogous. Then consider the reflections of a,b,c,d in bcd, acd, abd, abc, call them a',b',c',d'. Apply the condition to a'bcd, ab'cd, abc'd, abcd' and to the tetrahedron formed by their centers (its center is of course o). Also apply the condition to a'b'c'd' whose center is also o. We get a contradiction.

Consider two arbitrary points $A_1,B_1$ and choose $A_2,B_2$ inside $A_1B_1$ such that $A_1A_2=A_2B_2=B_2B_1$.Observe that there is $B,C,D$ such that $A_1BCD$ is a regular tetrahedron with incentre $A_2$ and $A_2BCD$ is a regular tetrahedron with centre $B_2$. So we have $f(A_2)=f(A_1)f(B)f(C)f(D)$ and $f(B_2)=f(B_1)f(B)f(C)f(D)$ from these 2 tetrahedrons.Deviding 2 equation we get $\frac{f(A_1)}{f(B_1)}=\frac{f(A_2)}{f(B_2)}$. Again consider $A_3,B_3$ inside $A_2,B_2$ in the same manner.In this way we get 2 infinite sequence of points $\{A_1,A_2,....\}$ and $\{B_1,B_2,...\}$ such that $\frac{f(A_i)}{f(B_i)}=\frac{f(A_{i+1})}{f(B_{i+1})}$.Note that $\{A_1,A_2,....\}$ and $\{B_1,B_2,...\}$ has same limit.So $\frac{f(A_1)}{f(B_1)}=\frac{f(A_2)}{f(B_2)}=\frac{f(A_3)}{f(B_3)}=\frac{f(A_4)}{f(B_4)}=....=1$.So $f(A_1)=f(A_2)$. So all the points are assigned with same number, let say by $r$ and from the 1st equation we have $r= r^4$.So r=1.

Solved with Elliott Liu, Jeffrey Chen, Luke Robitaille. For a tetrahedron \(ABCD\), let \(I\) be the incenter of \(ABCD\); let \(I_A\) be the incenter of \(IBCD\), with \(I_B\), \(I_C\), \(I_D\) defined similarly; let \(J_A\) be the incenter of \(I_ABCD\), with \(J_B\), \(J_C\), \(J_D\) defined similarly; and let \(J\) be the incenter of \(J_AJ_BJ_CJ_D\). Observe that \begin{align*} f(J)&=f(J_A)f(J_B)f(J_C)f(J_D)\\ &=f(I_A)f(I_B)f(I_C)f(I_D)\cdot(f(A)f(B)f(C)f(D))^3\\ &=f(I)^4\cdot(f(A)f(B)f(C)f(D))^6\\ &=f(I)^{10}. \end{align*} By reflecting the entire diagram through the midpoint of \(\overline{IJ}\), we have \(f(I)=f(J)^{10}\), so \(f(I)=f(I)^{100}\), implying \(f(I)=1\). This is sufficient. Remark: Note that we do not necessarily require \(I\ne J\).

Claim: Given $4$ collinear points $A,B,C,D$ in that order with $AB = BC = CD$, we have $$\frac{f(B)}{f(C)} = \frac{f(A)}{f(D)}$$ Proof: Let $X$ be the midpoint of $BC$. Construct points $P,Q,R$ on the plane through $X$ perpendicular to $BC$ such that $A$ is the center of the insphere of the tetrahedron $APQR$, so by symmetry $C$ is the center for $PQRD$. So $f(B) = f(A) f(P) f(Q) f(R)$ and $f(C) = f(D) f(P) f(Q) f(R)$, dividing, we get the desired result. $\square$ Let $X,Y$ be two arbitrary points, consider points $A,B,C$ on segment $XY$ such that $XA = AB = BC = CY$, so using the above claim on $XABC$ and $ABCY$, we get that $f(X) = f(Y)$, so $f$ is constant so if $f(X) = c$, then $c = c^4 \implies c = 1$, so $f(X) = 1$ for all points $X$, as desired. $\blacksquare$