Is the category of ordered fields thin? Pretty much everything I know about ordered fields, I learned in high school. So I apologize if this is a silly question.
The category of fields is not thin. For example, complex conjugation is a non-trivial automorphism of $\mathbb{C}$. What about the category of ordered fields? Presumably not, but an explicit counterexample would be nice.
 A: Well, any field homomorphism between fields of characteristic zero preserves $\mathbb{Q}$, so (by density), there is a unique field homomorphism between any two fields that contain $\mathbb{Q}$ as a dense subset (in the sense that between any two distinct elements there is a rational number). So we must look at ordered fields that are strictly bigger than $\mathbb{R}$.
Consider, for instance, a model $M$ of the theory obtained by adding to the theory of ordered fields two constants $\alpha$ and $\beta$ satisfying the following axioms:


*

*$\alpha \ne \beta$.

*$\varphi (\alpha) \leftrightarrow \varphi (\beta)$, for all well-formed formulae $\varphi$ in the language of ordered fields (i.e. not mentioning $\alpha$ or $\beta$) with one free variable.


Such a model $M$ exists, by compactness, and (the interpretations of) $\alpha$ and $\beta$ are distinct but indiscernible in $M$ (with respect to the language of ordered fields). $\mathbb{Q}$ is not dense in $M$ (because, if it were, then indiscernibility would imply $\alpha = \beta$), so there is at least a chance that $M$ has a non-trivial automorphism that exchanges $\alpha$ and $\beta$. However, in general this is only true after we pass to an elementary extension of $M$: see here. Nonetheless, we do obtain an ordered field equipped with a non-trivial automorphism.
