= Solution
A left <group action> is a map $(g,x)\mapsto g\cdot x$ with $e\cdot x=x$ and $(gh)\cdot x=g\cdot(h\cdot x)$. The <group orbit> and <stabiliser subgroup> are
$$
\operatorname{Orb}(x)=\{g\cdot x:g\in G\},\qquad G_x=\{g\in G:g\cdot x=x\}.
$$
The identity belongs to $G_x$. If $g,h\in G_x$, then $(gh)\cdot x=g\cdot x=x$; also $g^{-1}\cdot x=g^{-1}\cdot(g\cdot x)=x$. Thus $G_x$ is a <subgroup>.
For the <orbit-stabilizer theorem>, the map $gG_x\mapsto g\cdot x$ from left <cosets> to the <group orbit> is a bijection: $g\cdot x=h\cdot x$ if and only if $h^{-1}g\in G_x$, which is precisely $gG_x=hG_x$. Each left <coset> has $|G_x|$ elements, since multiplication by $g$ is a bijection from $G_x$ to $gG_x$, and the left <cosets> partition $G$. For finite $G$ this proves
$$
\boxed{|G|=|G_x|\,|\operatorname{Orb}(x)|.}
$$
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