Identity Mapping is Automorphism

Theorem
The identity mapping $I_S: \left({S, \circ}\right) \to \left({S, \circ}\right)$ on the algebraic structure $\left({S, \circ}\right)$ is an automorphism.

Its image is $S$.

Group Automorphism
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 $\left\{{e}\right\}$.

Ring Automorphism
Let $\left({R, +, \circ}\right)$ be a ring whose zero is $0$.

Then $I_R: \left({R, +, \circ}\right) \to \left({R, +, \circ}\right)$ is a ring automorphism.

Its kernel is $\left\{{0}\right\}$.

Proof
From the definition, an automorphism is an isomorphism from an algebraic structure onto itself.

An isomorphism, in turn, is a bijective homomorphism.

From Identity Mapping is a Bijection, the identity mapping $I_S: S \to S$ on the set $S$ is a bijection from $S$ onto itself.

Now we need to show it is a homomorphism.

Let $x, y \in S$. Then:

Thus $I_S \left({x \circ y}\right) = I_S \left({x}\right) \circ I_S \left({y}\right)$ and the morphism property holds, proving that $I_S: S \to S$ is a homomorphism.

As $I_S$ is a bijection, its image is $S$.

Proof for Group Automorphism
The main result holds directly.

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

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

Proof for Ring Automorphism
The main result holds directly, for both $+$ and $\circ$.

As $I_R$ is a bijection, the only element that maps to $0$ is $0$ itself.

Thus the kernel is $\left\{{0}\right\}$.