Could someone let me know if I did anything wrong in my answer?
Show that $Aut(G) \leq S_{G}$, the symmetric group.
Subset: Since $Aut(G)$ consists of functions that map elements from $G$ to $G$, these functions, which are bijective due to $Aut(G)$ being a set of isomorphisms, are permutations of the set $G$. Therefore, $Aut(G)\subset S_{G}$.
Identity: Both $Aut(G)$ and $S_{G}$ have an identity function that maps an element to itself. Since $Aut(G)\subset S_{G}$, it must be that that identity function is in $Aut(G)$.
Inverse: Since $Aut(G)$ is a set of isomorphisms, by definition, an inverse exists for each isomorphism.
Closure: Any bijective function composition with domain $G$ and codomain $G$ will be a bijective function with domain $G$ and codomain $G$. The operation is automatically preserved due to it being an isomorphism, and so the operation is closed.