Let $G$ be a finite $p$-group acting on a finite set $X$, and ${\rm Fix}_G(X)$ the subset of $X$ consisting of fixed points under this action. Then $$|X| \equiv | {\rm Fix}_G(X)| \pmod{p}.$$
The proof of this consists of saying $X$ is a disjoint union of orbits that all have a cardinality equal to $p^y$ where $y$ is an integer.
But surely this isn't true?
Using the $\Bbb Z_4$ group as an example, let $X = \{ 0,2,3\}$. There is one orbit here of length $4$, but $X$ does not have the element '$1$' in it, so it cannot be called a disjoint union of that orbit.
You seem to be assuming that if $G$ acts on a set $X$ and $Y$ is a subset of $X$, then the action restricts to an action of $G$ on $Y$. This is not the case in general; to make this work you need the hypothesis that $Y$ is what we call $G$-invariant, which means $g\in G$ and $y\in Y$ implies $gy\in Y$.
In the example you've given the set $\{0,2,3\}$ is not a $\mathbb Z_4$-invariant subset of $\{0,1,2,3\}$.