Solution:
The substitution $n = p$, a prime, yields $p \equiv (f(p))^{p} \equiv 0 \pmod{f(p)}$, so that $p$ is divisible by $f(p)$. Hence, for each prime $p$, $f(p) = 1$ or $f(p) = p$.

Let $S = \{p : p \text{ is prime and } f(p) = p\}$. If $S$ is infinite, then $f(n)^{p} \equiv n \pmod{p}$ for infinitely many primes $p$. By the little Fermat theorem, $n \equiv f(n)^{p} \equiv f(n)$, so that $f(n) - n$ is a multiple of $p$ for infinitely many primes $p$. This can happen only if $f(n) = n$ for all values of $n$, and it can be verified that this is a solution.

If $S$ is empty, then $f(p) = 1$ for all primes $p$, and any function satisfying this condition is a solution.

Now suppose that $S$ is finite and non-empty. Let $q$ be the largest prime in $S$. Suppose, if possible, that $q \geq 3$. Therefore, for any prime $p$ exceeding $q$, $p \equiv 1 \pmod{q}$. However, this is not true. Let $Q$ be the product of all the odd primes up to $q$. Then $Q + 2$ must have a prime factor exceeding $q$ and at least one of them must be incongruent to $1 \pmod{q}$. (An alternative argument notes that Bertrand's postulate can turn up a prime $p$ between $q$ and $2q$ which fails to satisfy $p \equiv 1 \pmod{q}$.)

The only remaining case is that $S = \{2\}$. Then $f(2) = 2$ and $f(p) = 1$ for every odd prime $p$. Since $f(n) 2 \equiv n \pmod{2}$, $f(n)$ and $n$ must have the same parity. Conversely, any function $f$ for which $f(n) \equiv n \pmod{2}$ for all $n$, $f(2) = 2$ and $f(p) = 1$ for all odd primes $p$ satisfies the condition.

Therefore the only solutions are
- $f(n) = n$ for all $n \in \mathbf{N}$;
- any function $f$ with $f(p) = 1$ for all primes $p$;
- any function for which $f(2) = 2$, $f(p) = 1$ for primes $p$ exceeding $2$ and $f(n)$ and $n$ have the same parity.