Let $n\in\mathbb{N}$ and $p$ be a prime. Consider $f_{p}:\mathbb{F}_{p}\rightarrow\mathbb{F}_{p}, x\mapsto x^{n}$. I want to show that $\liminf_{p\rightarrow\infty}\frac{\#Per(f_{p})}{p} = 0$, where $\#Per(f_{p})$ is the number of $f_{p}$-periodic points in $\mathbb{F}_{p}$.
My attempt: Notice that for every $r\geqslant 0$ the Dirichlet theorem for primes in arithmetic progressions implies that there exists infinitely many primes $p$ satisfying $p\equiv 1\mod n^{r}$. Notice that since $(\mathbb{F}_{p})^{*}$ is a cyclic group of order $p-1$, and thus since $n^{r}$ divides $p-1$ there are precisely $n^{r}$ elements in $(\mathbb{F}_{p})^{*}$ whose order divides $n^{r}$. One can show that a point $z$ in $(\mathbb{F}_{p})^{*}$ is $f_{p}$-periodic if and only if the order of $z$ is coprime to $n$. From here I am stuck.
You are most of the way there! You are clearly aware of the statement: if $d\mid(p-1)$, then there are exactly $d$ elements of $\Bbb F_p^*$ of order dividing $d$. Apply this with $d$ equal to the largest divisor $D$ of $p-1$ that is coprime to $n$, so that "order dividing $D$" is equivalent to "order coprime to $n$". It remains only to show that $D \mid (p-1)/n^r$ (under the assumption that $n^r\mid(p-1)$), which I offer to you as a challenge.