By the first condition we have $f(1)=f(1!)=f(1)!$ and $f(2)=f(2!)=f(2)!$ , so $f(1)=1$ or $2$ and similarly for $f(2)$ . By the second condition, we have $$n\cdot n!=(n+1)!-n! \mid f(n+1)!-f(n)! \qquad \qquad (1)$$ for all positive integers $n$ .
Suppose that for some $n \geq 2$ we have $f(n) = 1$ . We claim that $f(k)=1$ for all $k\ge n$ . Indeed, from Equation (1) we have $f(n+1)!\equiv 1 \mod n\cdot n!$ , and this is only possible if $f(n+1)=1$ ; the claim follows by induction.
We now divide into cases:
Case 1: $f(1)=f(2)=1$
This gives $f(n)=1$ always from the previous claim, which is a solution.
Case 2: $f(1)=2, f(2)=1$
This implies $f(n)=1$ for all $n\ge 2$ , but this does not satisfy the initial conditions. Indeed, we would have $$3-1 \mid f(3)-f(1)$$ and so $2\mid -1$ , a contradiction.
Case 3: $f(1)=1$ , $f(2)=2$
We claim $f(n)=n$ always by induction. The base cases are $n = 1$ and $n = 2$ . Fix $k > 1$ and suppose that $f(k)=k$ . By Equation (1) we have that $$f(k+1)! \equiv k! \mod k\cdot k! .$$ This implies $f(k+1)<2k$ (otherwise $f(k+1)!\equiv 0 \mod k\cdot k!$ ). Also we have $$(k+1)-1 \mid f(k+1)-f(1)$$ so $f(k+1)\equiv 1 \mod k$ . This gives the solutions $f(k+1)=1$ and $f(k+1)=k+1$ . The first case is obviously impossible, so $f(k + 1) = k + 1$ , as desired. By induction, $f(n) = n$ for all $n$ . This also satisfies the requirements.
Case 4: $f(1)=f(2)=2$
We claim $f(n)=2$ by a similar induction. Again if $f(k)=2$ , then by (1) we have $$f(k+1)\equiv 2 \mod k\cdot k!$$ and so $f(k+1)<2k$ . Also note that $$k+1-1 \mid f(k+1)-2$$ and $$k+1-2 \mid f(k+1)-2$$ so $f(k+1)\equiv 2 \mod k(k-1)$ . Then the only possible solution is $f(k+1)=2$ . By induction, $f(n) = 2$ for all $n$ , and this satisfies all requirements.
In summary, there are three solutions: $\boxed{f(n)=1, f(n)=2, f(n)=n}$ .