If $G$ is $p$-group acting on a finite set $X$, prove that $|X| \equiv |\mathrm{Fix}_G(X)|\pmod p$.
I've proved it when $G$ is a finite group using the equation $|X| = |\mathrm{Fix}_G(X)| + \sum |O(x_i)|$ and the fact that $|G| = p^k$. In that situation as $|O(x_i)|$ $|$ $|G|$ then $|O(x_i)|$ is a multiple of p so that taking congruences on both sides I get the result.
Is this true when $G$ is infinite? Please give me a hint of what I should use or a counterexample.