Automorphism Group is Subgroup of Symmetric Group

Theorem
Let $\struct {S, *}$ be an algebraic structure.

Let $\Aut S$ be the automorphism group of $\struct {S, *}$.

Then $\Aut S$ is a subgroup of the symmetric group $\struct {\Gamma \paren S, \circ}$ on $S$.

Proof
An automorphism is an isomorphism $\phi: S \to S$ from an algebraic structure $S$ to itself.

The Identity Mapping is Automorphism, so $\Aut S$ is not empty.

The composite of isomorphisms is itself an isomorphism, as demonstrated on Isomorphism is Equivalence Relation.

So:
 * $\phi_1, \phi_2 \in \Aut S \implies \phi_1 \circ \phi_2 \in \Aut S$

demonstrating closure.

If $\phi \in \Aut S$, then $\phi$ is bijective and an isomorphism.

Hence from Inverse of Algebraic Structure Isomorphism is Isomorphism, $\phi^{-1}$ is also bijective and an isomorphism.

So $\phi^{-1} \in \Aut S$.

The result follows by the Two-Step Subgroup Test.