Conjugacy Action on Identity

Theorem
Let $G$ be a group whose identity is $e$.

For the conjugacy action:
 * $\left|{\operatorname{Orb} \left({e}\right)}\right| = 1$

and thus:
 * $\operatorname{Stab} \left({e}\right) = G$

Proof
So the only conjugate of $e$ is $e$ itself.

Thus:
 * $\operatorname{Orb} \left({e}\right) = \left\{{e}\right\}$

and so:
 * $\left|{\operatorname{Orb} \left({e}\right)}\right| = 1$

From the Orbit-Stabilizer Theorem, it follows immediately that:
 * $\operatorname{Stab} \left({e}\right) = G$