Great question - surprisingly, the ordering on $\mathbb{R}$ is definable purely algebraically! We have $$x\le y\quad\iff\quad \exists z(z^2+x=y).$$ This, in combination with the fact that any field automorphism fixes $\mathbb{Q}$ pointwise, shows that the only field automorphism of $\mathbb{R}$ is the identity.
(Note that this suggests a way to get subfields of $\mathbb{R}$ which do have nontrivial automorphisms: whip up a subfield with some crucial elements missing square roots. And this does work: e.g. in $\mathbb{Q}(\pi)$ we can get a nontrivial automorphism by swapping $\pi$ and $-\pi$, roughly because the fact that $\sqrt{\pi}\not\in\mathbb{Q}(\pi)$ prevents us from distinguishing $\pi$ and $-\pi$)
Note that this breaks down when we move to $\mathbb{C}$, since every element has a square root. Indeed, $\mathbb{C}$ has an obvious nice nontrivial automorphism (conjugation) and assuming the axiom of choice has lots of wild nontrivial automorphisms. (If we drop choice it is consistent that the only automorphisms are the identity and conjugation.)