Automorphism Group is Subgroup of Symmetric Group

Theorem
Let $\left({S, *}\right)$ be an algebraic structure.

Let $\operatorname{Aut} \left({S}\right)$ be the group of automorphisms of $\left({S, *}\right)$.

Then $\operatorname{Aut} \left({S}\right)$ is a subgroup of the group of permutations $\left({\Gamma \left({S}\right), \circ}\right)$ on the underlying set $S$ of $\left({S, *}\right)$.

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

The Identity Mapping is Automorphism, so $\operatorname{Aut} \left({S}\right)$ is not empty.

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

So:
 * $\phi_1, \phi_2 \in \operatorname{Aut} \left({S}\right) \implies \phi_1 \circ \phi_2 \in \operatorname{Aut} \left({S}\right)$

demonstrating closure.

If $\phi \in \operatorname{Aut} \left({S}\right)$, 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 $\phi^{-1} \in \operatorname{Aut} \left({S}\right)$.

The result follows by the Two-Step Subgroup Test.