Showing $\operatorname{Aut}(\mathbb{R})$ is abelian I have an exercise (not for a class) that asks whether $\operatorname{Aut}(\mathbb{R})$ (field automorphisms) is abelian.  I know that $\operatorname{Aut}(\mathbb{R})$ is just the trivial group, but is there a nice way to see that it is abelian without knowing the group?

This is Part (c) of a question and Parts (a) and (b) were that $\operatorname{Aut}(\overline{\mathbb{Q}})$ is infinite and nonabelian.
$\mathbb R$ is a $\mathbb Q$-vector space, so every $\mathbb Q$-linear function is in $Aut(\mathbb R)$ and $End_{\mathbb Q}(\mathbb R)$ isn't abelian, so $Aut(\mathbb R)$ isn't abelian.

If you mean $Aut(\mathbb R)$ like the group of ring automorphism of $\mathbb R$, the only one is the identity.
$\mathbb R$ is a $\mathbb Q$-vector space, so every $\mathbb Q$-linear function is in $Aut(\mathbb R)$ and $End_{\mathbb Q}(\mathbb R)$ isn't abelian, so $Aut(\mathbb R)$ isn't abelian.

If you mean $Aut(\mathbb R)$ like the group of ring automorphism of $\mathbb R$, the only one is the identity.