Proof that stabilisers of points in the same orbit are conjugate

Proof

We prove the two inclusions between $G_{g \cdot x}$ and $g G_x g^{-1}$. Let $e$ be the identity of $G$, and let $y = g \cdot x$.

1. Let $h \in G$. By the axioms of an action, $(g^{-1} h g) \cdot x = g^{-1} \cdot (h \cdot (g \cdot x)) = g^{-1} \cdot (h \cdot y)$.
2. Suppose $h \in G_y$, the stabiliser of $y$. Then step 1 gives $(g^{-1} h g) \cdot x = g^{-1} \cdot y = g^{-1} \cdot (g \cdot x) = e \cdot x = x$.
3. So $g^{-1} h g \in G_x$, and $h = g (g^{-1} h g) g^{-1} \in g G_x g^{-1}$. Thus $G_y \subseteq g G_x g^{-1}$.
4. Conversely, let $k \in G_x$ and $h = g k g^{-1}$. Then $h \cdot y = (g k g^{-1}) \cdot (g \cdot x) = (g k) \cdot x = g \cdot (k \cdot x) = g \cdot x = y$.
5. So $h \in G_y$, and $g G_x g^{-1} \subseteq G_y$.
6. By steps 3 and 5, $G_{g \cdot x} = g G_x g^{-1}$, a conjugate subgroup of $G_x$.