Show that, for each finite $G$-set ($G$-action) on $X$, we have $|X| \equiv |X^G| \pmod n$

Question

Let $G$ a group and $n \in \mathbb{Z}_{>0}$ an integers with the following properties. For each subgroup $H < G$ such that $H \not= G$, the integer $n \mid [G:H]$. Show that, for each finite $G$-set ($G$-action) on $X$, we have $|X| \equiv |X^G| \pmod n$

I think I can suppose $|X^G| \geq n$, and prove there exists $H < G$ , $H \ne G$, the integer $n \nmid [G:H]$.

I'm on this problem for a while now. Does someone could give me a little hint?

Answer 1

Here is a hint: show that $n$ divides $|X \setminus X^G|$. To show this, let $G$ act on $X$ and show that every orbit has length $1$ or a multiple of $n$.

The following spoiler contains another hint:

Assume an orbit has length $x>1$ and apply the orbit-stabilizer formula.

Answer 2

Let $X$ be a $G$ set, $X$ is the disjoint of the orbits of $G$. Let $x\in X$ the cardinal of orbit $Ob(x)$ of $x$ is in bijection with $|G|/|G_x|=[G:G_x]$ where $G_x$ is the stabilizer of $x$, we deduce that $n$ divides the cardinal of $Orb(x)$ if $x$ is not fixed by $G$. $|X|=|X^G|+\sum_{Orb(x), x \neq X^G}|Orb(x)|$, since $|Orb(x)|=0 mod$ $n$ if $x\neq X^G$, we deduce that $|X|=|X|^G mod$ $n$.