= Solution
For $x\in X$,
$$
\operatorname{Orb}(x)=\{gx:g\in G\},
\qquad
\operatorname{Stab}(x)=\{g\in G:gx=x\}.
$$
The stabilizer contains the identity; if $g,h$ fix $x$, then
$$
(gh^{-1})x=g(h^{-1}x)=gx=x,
$$
so it is a subgroup.
The <orbit-stabilizer theorem> states
$$
\boxed{|G|=|\operatorname{Orb}(x)|\,|\operatorname{Stab}(x)|}.
$$
Define
$$
G/\operatorname{Stab}(x)\longrightarrow\operatorname{Orb}(x),
\qquad
g\operatorname{Stab}(x)\longmapsto gx.
$$
The map is well defined and bijective: two elements give the same image exactly when they differ by an element of the stabilizer. Lagrange's theorem then proves the formula.
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