Definition:Automorphism Group/Group

Theorem
The set of automorphisms of an algebraic structure $\left({S, \circ}\right)$ is a group, under composition of mappings, and is denoted $\operatorname{Aut} \left({S}\right)$.

It is a subgroup of the Group of Permutations $\left({\Gamma \left({S}\right), \circ}\right)$ on the underlying set of $\left({S, \circ}\right)$.

The structure $\left({S, \circ}\right)$ is usually a group. However, this is not necessary for this result to hold.

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


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


 * The composite of isomorphisms is itself an isomorphism, as demonstrated here.

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 $\phi \in \operatorname{Aut} \left({G}\right)$, then $\phi$ is bijective and an isomorphism.

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

So $\phi^{-1} \in \operatorname{Aut} \left({G}\right)$.

The result follows by the Two-Step Subgroup Test.