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 $\left|{\operatorname{Orb} \left({e}\right)}\right| = 1$.

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