So I remember reading once that the only element of $G=Gal(\overline{\Bbb Q} / \Bbb Q)$ that we understand is complex conjugation. Suppose we fix an embedding of $\overline{\Bbb Q}$ into $\Bbb C$. Then complex conjugation in $\Bbb C$ restricts to a "complex conjugation", $\sigma \in G$. Clearly $\sigma$ has order $2$, and any conjugate of $\sigma$ also has order $2$.
Question: Is it true that any element of order $2$ in $G$ is conjugate to $\sigma$?
I've thought about this for a while, but my knowledge of elements of $G$ isn't excellent. For general groups, not all elements of order $2$ are necessarily conjugate. For example, $(12)$ and $(12)(34)$ are both of order $2$ but not conjugate in $S_n$ for any $n\geq 4$. It is unclear how to proceed in the case of $G$.
My reason for asking is that the element $\sigma$ I defined above should only be thought of as being defined up to conjugacy, since a different embedding gives a new complex conjugation which is conjugate to $\sigma$. Thus, if all elements of order $2$ in $G$ were conjugate then I believe I could conclude that there is really only a single honest complex conjugation.
I apologize if any of this is unclear, I'd be glad to elaborate if necessary. Also, I'd appreciate any intuition on how to think about this in terms of Galois representations, since those seem to be the only way to get a handle on $G$.
It is a theorem of E. Artin that any non-trivial finite order element of $\mathbb Q$ is a complex conjugation. This follows from the theory of real closed fields; see here for example. (If $H \subset G$ is a non-trivial finite subgroup, then $F:=\overline{\mathbb Q}^H$ is a field whose algebraic closure is a proper finite exension; thus $F$ satisfies property 4 from the linked page. Property 5 then shows that $\overline{\mathbb Q} = F(\sqrt{-1})$, from which one sees that $H$ has order $2$; a little more argument shows that the non-trivial element of $H$ acts as a complex conjugation. Any treatment of real closed fields that is more detailed than the wikipedia page should give the details.)