Let $\overline{\mathbb{Q}}$ be the algebraic closure of $\mathbb{Q}$. The absolute galois group $G\_\mathbb{Q}$ of $\mathbb{Q}$ acts on the set of real-closed subfields of $\overline{\mathbb{Q}}$.



> 
> Does it act transitively?
> 
> 
> 


The real-closed subfields are in bijection with the involutions of $G\_\mathbb{Q}$ under the Galois correspondence, so another way to ask the question would be,



> 
> Do the involutions of $G\_{\mathbb{Q}}$ form a single conjugacy class?
> 
> 
> 


I am asking out of curiosity. I have been unable to locate the answer in any of my texts on real fields, or via internet search, but please forgive me if it is well known. Because the order on a real-closed field is unique, and $\mathbb{Q}$ is order-dense in its real closure, the real-closed subfields $K$ of $\overline{\mathbb{Q}}$ have trivial automorphism group, and it follows that the stabilizer of each $K$ for the action of $G\_\mathbb{Q}$ is just the involution fixing $K$ pointwise; thus the action is almost free, and this made me curious if it is transitive.



Are all real-closed subfields of $\overline{\mathbb{Q}}$ conjugate?
Any real-closed subfield $R\subseteq\overline{\mathbb Q}$ is a real closure of $\mathbb Q$ (being real closed and algebraic over $\mathbb Q$). Thus, by uniqueness of real closures, any two such fields are isomorphic, and an isomorphism of $R$ to $R'$ extends to an isomorphism of $R(i)=\overline{\mathbb Q}$ to $R'(i)=\overline{\mathbb Q}$, i.e., an automorphism of $\overline{\mathbb Q}$.


