Stabilizer is Normal iff Stabilizer of Each Element of Orbit

Theorem
Let $\struct {G, \circ}$ be a group.

Let $S$ be a set.

Let $*: G \times S \to S$ be a group action.

Let $x \in S$.

Let $\Stab x$ denote the stabilizer of $x$ under $*$.

Let $\Orb x$ denote the orbit of $x$ under $*$.

Then $\Stab x$ is normal in $G$ $\Stab x$ is also the stabilizer of every element in $\Orb x$.

Necessary Condition
Let $\Stab x$ be normal in $G$.

Let $t \in \Stab x$.

Let $y \in \Orb x$.

By :
 * $(1): \quad y = g * x$

for a $g \in G$.

By :
 * ${\Stab x} \circ g = g \circ \Stab x$

In particular:
 * $(2): \quad t \circ g = g \circ t'$

for a $t' \in \Stab x$.

Therefore:

That is:
 * $t \in \Stab y$

Sufficient Condition
Let $\Stab x$ be the stabilizer of every element in $\Orb x$.

We shall show, i.e.
 * $\forall g \in G : g^{-1} \circ \Stab x \circ g \subseteq \Stab x$
 * $\forall g \in G : g \circ \Stab x \circ g^{-1} \subseteq \Stab x$

It suffices to verify the first inclusion, from which the second one follows by choosing $g^{-1}$ instead of $g$.

To this end, let $g \in G$.

Let $t \in \Stab x$.

By :
 * $g * x \in \Orb x$

Thus :
 * $(3): \quad t \in \Stab {g * x}$

Therefore:

That is:
 * $g^{-1} \circ t \circ g \in \Stab x$