Identity Mapping is Automorphism

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

Its image is $S$.

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

An isomorphism, in turn, is a bijective homomorphism.

From Identity Mapping is 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 $\map {I_S} {x \circ y} = \map {I_S} x \circ \map {I_S} y$ and the morphism property holds/

This proves that $I_S: S \to S$ is a homomorphism.

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