Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that $f(n!)=f(n)!$ for all positive integers $n$ and such that $m-n$ divides $f(m)-f(n)$ for all distinct positive integers $m, n$.
Answer: $f \equiv 1, f \equiv 2$, and $f$ the identity. As these obviously work, we prove these are the only ones.
By putting $n=1$ and $n=2$ we give $f(1), f(2) \in\{1,2\}$. Also, we will use the condition $m!-n!$ divides $f(m)!-f(n)!$.

We consider four cases on $f(1)$ and $f(2)$, and dispense with three of them.
- If $f(2)=1$ then for all $m \geq 3$ we have $m!-2$ divides $f(m)!-1$, so $f(m)=1$ for modulo 2 reasons. Then clearly $f(1)=1$.
- If $f(1)=f(2)=2$ we first obtain $3!-1 \mid f(3)!-2$, which implies $f(3)=2$. Then $m!-3 \mid f(m)!-2$ for $m \geq 4$ implies $f(m)=2$ for modulo 3 reasons.

Hence we are left with the case where $f(1)=1$ and $f(2)=2$. Continuing, we have
$$
3!-1 \mid f(3)!-1 \quad \text { and } \quad 3!-2 \mid f(3)!-2 \Longrightarrow f(3)=3 \text {. }
$$

Continuing by induction, suppose $f(1)=1, \ldots, f(k)=k$.
$$
k!\cdot k=(k+1)!-k!\mid f(k+1)!-k!
$$
and thus we deduce that $f(k+1) \geq k$, and hence
$$
k \left\lvert\, \frac{f(k+1)!}{k!}-1\right.
$$

Then plainly $f(k+1) \leq 2 k$ for $\bmod k$ reasons, but also $f(k+1) \equiv 1(\bmod k)$ so we conclude $f(k+1)=k+1$.
Remark. Shankar Padmanabhan gives the following way to finish after verifying that $f(3)=3$. Note that if
$$
M=((((3!)!)!)!\ldots)!
$$
for any number of iterated factorials then $f(M)=M$. Thus for any $n$, we have
$$
M-n|f(M)-f(n)=M-f(n) \Longrightarrow M-n| n-f(n)
$$
and so taking $M$ large enough implies $f(n)=n$.