I got the following problem, and I had no clue to prove it. Could someone help me?
The problem is:
Let $G$ act transitively on a set $X$. Fix $x_{0} \in X$, let $H$ = Stab($x_0)$, and let $Y$ denote the set of conjugates of $H$ in $G$. Show that there is a $G$-equivariant surjective map from $X$ to $Y$ given by $x \to \text{Stab}(x)$, and this map is $[N_{G}(H) : H]$ to one.
Let $x\in X$, since $G$ acts transitively, there exists $g\in G$ such that $g.x_0=x$, $Stab(x)=gHg^{-1}$ this implies that $f:x\rightarrow Stab(x)$ takes it values in $Y$ and is equivariant $(f(g.x)=Stab(g.x)=gStab(x)g^{-1}=gf(x)g^{-1})$.
$f$ is surjective,for every $gHg^{-1}\in Y$, $f(g.x_0)=gHg^{-1}$.
Suppose that $f(x)=f(x_0)$ write $x=gx_0$, $gHg^{-1}=H$ implies that $g\in N_G(H)$ the fibre of $x_0$ by $f$ is the cardinal of $N_G(H).x_0$, the orbit of $x_0$ by $N_G(H)$, the kernel of the restriction of $f$ to $N_G(H)$ is $H$, we deduce that $|N_G(H)|.x_0=|N_G(H)/H|$.