Let $G$ be a group and let $X$ be a finite $G$-set. Assume that $g\in G$ is of finite order $p^n$, where $p$ is a prime number and $n\ge 1$. Let $h=g^{p^{n-1}}$. Show that $|X|=|X^h|\pmod {p^n}$, where $X^h=\{x\in X: h\cdot x=x\}$.
Using the identity $$ |X|=|X^h|+\sum_i [\langle h \rangle : \text{Stab}_{\langle h \rangle}(x_i)] $$ where no $x_i\in X$ is fixed by all elements in $\langle h \rangle$, we derive that $|X|=|X^h|+p\cdot \#\text{non-trivial $\langle h \rangle$-orbits}$. It remains to show that the number of non-trivial $\langle h \rangle$-orbits is divisible by $p^{n-1}$. I struggle to proceed. If we apply Burnside's lemma, we then need to show that $p^n$ divides $\sum_{m=0}^{p-1} |X^{h^m}|$, which I have no idea about. Am I missing something?
Consider the action of $g$. Since $\langle h\rangle$ is contained in any nontrivial subgroup of $\langle g\rangle$, $h$ is contained in the stabilizer of any $x\in X$ whose orbit has fewer than $p^n$ elements. Thus, all those elements are in $X^h$. The remaining elements lie in orbits of size $p^n$, which get partitioned into orbits of size $p$ by $h$; these are the terms in the sum, which is therefore a multiple of $p^n$.