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.

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.