Is an automorphism of the field of real numbers without ordering the identity map? [duplicate]

We know an automorphism on $$\mathbb{R}$$ must fix $$\mathbb{Q}$$. If we assume the usual order structure and topology on $$\mathbb{R}$$, then we can use the density of $$\mathbb{Q}$$ to show an automorphism must be the identity map, similar to the answers in this thread.

If we do not assume the usual ordering on $$\mathbb{R}$$, is it still true that any automorphism is the identity map?

• You don't seem to have read the question you linked very carefully. It is asking exactly the question you have asked, without assuming anything about preserving the order and topology. – Eric Wofsey Dec 13 '19 at 21:44

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.)