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.