Identity Mapping is Automorphism/Groups

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Then $I_G: \left({G, \circ}\right) \to \left({G, \circ}\right)$ is a group automorphism.

Its kernel is $\left\{{e}\right\}$.

Proof
The main result Identity Mapping is Automorphism holds directly.

As $I_G$ is a bijection, the only element that maps to $e$ is $e$ itself.

Thus the kernel is $\left\{{e}\right\}$.