This is a very narrow question I haven't been able to find answered elsewhere.
Given a group $G$, I want to prove that $\mathrm{Aut}(G)$ is a group.
All of the steps make sense to me, but I'm wondering about proving closure. For any $f,g \in \mathrm{Aut}(G)$, so I need to prove both that $f \circ g$ and $g \circ f$ are elements of $\mathrm{Aut}(G)$? What qualifies as closure in this case?
When I normally prove that a group is closed, I take arbitrary $a,b$ and prove that $ab \in G$. Even then I'm questioning why I don't need that $ba$ is also an element of $G$. Is this drawing on symmetry of the labels?
Automorphisms are bijections and the composition of bijections is bijections.
So we only need to show compatibility with the group operation.
If $x,y \in G$, $f,g \in {\rm Aut}(G)$,
$$\begin{align} (f\circ g)(xy) &= f(g(xy)) \\ &= f(g(x)g(y)) \\ &= f(g(x))(f(g(y)) \\ &= (f\circ g)(x)(f\circ g)(y). \end{align}$$
So we can see $f\circ g$ is a bijection from $G$ to itself and respects the group operation, hence an automorphism.