Lemma 1: $x=d(x+1),(*) \leftrightarrow x \in \left\{2;3 \right\}.$ Proof: $\,$ Let $x+1=2^a.3^b.p_1^{e_1}...p_k^{e_k}.$ $\,$ Then $(*) \leftrightarrow 2^a.3^b.p_1^{e_1}...p_k^{e_k}=(a+1)(b+1)(e_1+1)(e_2+1)...(e_k+1)+1.$ $\,$ Note that, for any $ p\geq 5,$ we have: $p^k \geq 4k+1;$ $3^a \geq 2a+1$ and $2^a \geq max \left\{ 2a;a+1 \right\}.$ $\,$ If $x+1$ has a prime divisor $\geq 5,$ or $e_1 \geq 1,$ then \begin{align*}x+1 & \geq (a+1)(2b+1)(4e_1+1)...(4e_k+1) \\ & =(a+1)(b+b+1)(3e_1+e_1+1)...(4e_k+1) \\ & \geq (a+1)(b+1)(e_1+1)...(e_k+1)+3e_1 \\ & \geq (a+1)(b+1)(e_1+1)...(e_k+1)+3>d(x+1)+1, \end{align*}$\,$ adsurb. $\,$ So $x+1=2^a.3^b,$ $(a+1)(b+1)+1=d(x+1)+1=x+1=2^a3^b.$ $$ \Rightarrow (a+1)(b+1)+1 \geq 2a(2b+1)+1 $$$\,$ $ \leftrightarrow 2 \geq 3ab+a.$ Thus, $\left\{a;b \right\}=\left\{2;0 \right\}, \left\{0;1 \right\}.$ Or $x \in \left\{2;3 \right\}.$ q.e.d $$\left\{\begin{matrix}f(d(n+1))=d(f(n)+1), (1) & & \\f\left ( \sigma (n+1) \right )=\sigma \left ( f(n)+1 \right ), (2) & & \\ \end{matrix}\right.$$$\bullet$ Let $n=3$ in $(1),$ from Claim 1, we get: $f(3)\in \left\{2;3 \right\}.$ $\bullet$ Let $n=1$ in $(2),$ we get: $\sigma\left ( f(1)+1 \right ) =f(3) \in \left\{2;3 \right\}.$ But there is no positive $x$ such that $\sigma(x)=2 \rightarrow f(3)=3 \rightarrow \sigma\left ( f(1)+1 \right ) =3 \rightarrow f(1)=1.$ Claim 1: $f(2^k-1)=2^k-1, \forall k.$ Proof: $\,$ I'll prove this by induction, the case $n=1,2$ is already proved. $\,$ If $f(2^k-1)=2^k-1,$ let $n \rightarrow 2^k-1$ in $(2),$ we get: $$f \left( \sigma \left(2^k \right) \right)= \sigma \left( f\left(2^k-1 \right)+1 \right)= \sigma \left( 2^k \right) \leftrightarrow f \left( 2^{k+1}-1 \right)= 2^{k+1}-1 .$$q.e.d $\bullet$ For any $k,$ let $x \rightarrow 2^k -1$ in $(1),$ we get: $$f\left ( d\left ( 2^k \right ) \right )=d\left ( f\left ( 2^k-1 \right ) +1\right )=d\left ( 2^k \right )\leftrightarrow f(k+1)=k+1.$$Thus, $$f(n)=n, \forall n.$$ Q.E.D