There are three such functions: the constant functions 1, 2 and the identity function $\mathrm{id}_{\mathbf{Z}^{+}}$. These functions clearly satisfy the conditions in the hypothesis. Let us prove that there are only ones.

Consider such a function $f$ and suppose that it has a fixed point $a \geq 3$, that is $f(a)=a$. Then $a!,(a!)!, \cdots$ are all fixed points of $f$, hence the function $f$ has a strictly increasing sequence $a_{1}<a_{2}<\cdots<a_{k}<\cdots$ of fixed points. For a positive integer $n$, $a_{k}-n$ divides $a_{k}-f(n)=$ $f\left(a_{k}\right)-f(n)$ for every $k \in \mathbf{Z}^{+}$. Also $a_{k}-n$ divides $a_{k}-n$, so it divides $a_{k}-f(n)-\left(a_{k}-n\right)=$ $n-f(n)$. This is possible only if $f(n)=n$, hence in this case we get $f=\mathrm{id}_{\mathbf{Z}^{+}}$.

Now suppose that $f$ has no fixed points greater than 2. Let $p \geq 5$ be a prime and notice that by Wilson's Theorem we have $(p-2)!\equiv 1(\bmod p)$. Therefore $p$ divides $(p-2)!-1$. But $(p-2)!-1$ divides $f((p-2)!)-f(1)$, hence $p$ divides $f((p-2)!)-f(1)=(f(p-2))!-f(1)$. Clearly we have $f(1)=1$ or $f(1)=2$. As $p \geq 5$, the fact that $p$ divides $(f(p-2))!-f(1)$ implies that $f(p-2)<p$. It is easy to check, again by Wilson's Theorem, that $p$ does not divide $(p-1)!-1$ and $(p-1)!-2$, hence we deduce that $f(p-2) \leq p-2$. On the other hand, $p-3=(p-2)-1$ divides $f(p-2)-f(1) \leq(p-2)-1$. Thus either $f(p-2)=f(1)$ or $f(p-2)=p-2$. As $p-2 \geq 3$, the last case is excluded, since the function $f$ has no fixed points greater than 2. It follows $f(p-2)=f(1)$ and this property holds for all primes $p \geq 5$. Taking $n$ any positive integer, we deduce that $p-2-n$ divides $f(p-2)-f(n)=f(1)-f(n)$ for all primes $p \geq 5$. Thus $f(n)=f(1)$, hence $f$ is the constant function 1 or 2.