Let $q\neq p$ be a prime and let $g$ be a primitive root modulo $q$. Then $g^{a}\equiv p\pmod q$ for some positive integer $a$. We now see that $n^{p}\equiv p\pmod q$ has no solution in $n$ if and only if $bp\equiv a\pmod{(q-1)}$ has no solution $b$, since there must be a $b$ so that $g^{b}\equiv n\pmod q$. However, $bp\equiv a \pmod{(q-1)}$ certainly has a solution in $b$ if $p$ has an multiplicative inverse modulo $q-1$, ie if $\gcd(p,q-1)=1$, so $p\mid q-1$ is a necessary condition. Now, if $n^{p}\equiv p\pmod q$ for some $n$, then it follows that \[p^{\frac{q-1}{p}}\equiv n^{p\cdot\frac{q-1}{p}}\equiv n^{q-1}\equiv 1\pmod q,\] so let's prove that there exist a prime so that $q\equiv 1 \pmod p$ $p^{\frac{q-1}{p}}\not\equiv 1\pmod q$, that is, $ord_{q}(p)\nmid \frac{q-1}{p}$. We shall now prove that there exist a prime $q$ which satisfies $q\equiv 1\pmod p$ $ord_{q}(p)=p$ $p\nmid \frac{q-1}{p}$, that is, $p^{2}\nmid q-1$ Since $p$ is prime, we have $ord_{q}(p)=p$ if and only if $p^{p}\equiv 1\pmod q$ $p^{1}\not\equiv 1\pmod q$ which is equivalent to \[p^{p-1}+\ldots+p^{2}+p+1\equiv 0 \pmod q.\] Let $q_{1}\cdot\ldots\cdot q_{l}= p^{p-1}+\ldots+p^{2}+p+1$, where $q_{i}$ are prime numbers. By construction, every $q_{i}$ satisfies $p\mid q_{i}-1$, however $q_{i}\equiv 1\pmod{p^{2}}$ cannot hold for every $q_{i}$ since this would imply $p+1\equiv 1\pmod{p^{2}}$, a contradiction. Hence, we have found a $q$ which satisfies all conditions. $\Box$

At first consider that: \[\frac{p^p-1}{p-1}\equiv{p^{p-1}+p^{p-2}+\cdots+p^2+p+1}\equiv{p+1}\not\equiv{0}~~~\mbox{(mod p^2)}\] Now if all of the prime factors of $\frac{p^p-1}{p-1}$ like $q$ be equivalent $1$ modulo $p^2$, then we have: \[ \frac{p^p-1}{p-1}\equiv {1}\not\equiv{p+1}~~~\mbox{(mod p^2)}\] Which is a contradiction. So there exists a prime $q$ such that $q|\frac{p^p-1}{p-1}$ and $q\not\equiv{1}~~\mbox{(mod p^2)}$ We claim that this $q$ convince all of the properties of the problem. Now we check two cases: Case 1: \[q|p-1~~\Longrightarrow~~{p}\equiv{1}~~\Longrightarrow~~{p^p}\equiv{1}~~~\mbox{(mod q)}\] Case 1: \[{q}\nmid{p-1}~~\Longrightarrow~~gcd(p-1,q)=1~~\Longrightarrow~~{p^p-1}\equiv{0}~~\Longrightarrow~~{p^p}\equiv{1}~~~\mbox{(mod q)}\] So we always have $p^p\equiv{0}~~\mbox{(mod q)}$. Now suppose that $q|n^p-p$, which means: \[{n^p}\equiv{p}~~\Longrightarrow~~{n^{p^2}}\equiv{p^p}\equiv{1}~~~\mbox{(mod q)}\] Now let $ord_{q}{(n)}=d$. We have: \[d|p^2~~\Longrightarrow~~\left\{\begin{array}{cc}{a)}d=1\\{b)}d=p\\{c)}d=p^2\end{array}\right.\] If $d=1$ then we have $n\equiv{1}~~\mbox{(mod q)}$, so: \[1\equiv{n^p}\equiv{p}~~~\mbox{(mod q)}~~\Longrightarrow~~{p^{p-1}+p^{p-2}+\cdots+p^2+p+1}\equiv{p}~~~\mbox{(mod q)}\] But we have: \[\frac{p^p-1}{p-1}\equiv{p^{p-1}+p^{p-2}+\cdots+p^2+p+1}\equiv{0}~~~\mbox{(mod q)}\] So we should have $p\equiv0~~\mbox{(mod q)}$ which is a contradiction. If $d=p$ then we have $1\equiv{n^p}\equiv{p}~~\mbox{(mod q)}$ which follows the previous case. And at last if $d=p^2$, it means that $p^2|\varphi{(q)}=q-1$ which is a contradiction since we supposed that $q\not\equiv{1}~~\mbox{(mod p^2)}$. So we are done.

PS

Thank you all for the problem and answers. Is there some example of the using this result?