Keenan Kidwell's answer to [Place stabilizers for the absolute Galois Group](https://mathoverflow.net/questions/107052/place-stabilizers-for-the-absolute-galois-group/107057#107057) mentions that "choosing a complex conjugation" in $G\_{\mathbf{Q}}$ means choosing an embedding $\overline{\mathbf{Q}}\rightarrow\mathbf{C}$ so the consequent injection $\mathrm{Gal}(\mathbf{C}/\mathbf{R})\hookrightarrow G\_{\mathbf{Q}}$ takes ordinary complex conjugation to an element of $G\_{\mathbf{Q}}$. Obviously such an element has order 2. Is there some natural group theoretic characterization of exactly which elements of $G\_{\mathbf{Q}}$ these can be? 



Embeddings of $\overline{\mathbf{Q}}$ into $\mathbf{C}$
$\def\QQ{\mathbb{Q}}\def\RR{\mathbb{R}}$A community wiki answer to record the proof sketched above. I had misread the question earlier; the comments of S.Carnahan and user36938 are correct.


Let $\sigma$ be an element of order $2$ in $Gal(\bar{\QQ}/\QQ)$ and let $R$ be the fixed field of $\sigma$. By the [Artin-Schrier theorem](http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/artinschreier.pdf), $R$ is [real closed](https://en.wikipedia.org/wiki/Real_closed_field). In particular, it comes with a natural order $\leq\_R$ defined by $a \leq\_R b$ if $\sqrt{b-a} \in R$.


**Lemma** The order $\leq\_R$ is archimedean. 


**Proof** Suppose, for the sake of contradiction, that there is some $t \in R$ with $t >\_R a$ for all $a \in \QQ$. Let $t^n + a\_{n-1} t^{n-1} \cdots + a\_0$ be the minimal polynomial of $t$ over $\QQ$. But then 
$$|a\_{n-1} t^{n-1} \cdots + a\_0| \leq\_R t^{n-1} \sum | a\_i | <\_R t^n,$$
a contradiction. Here, for $x \in R$, the notation $|x|$ means whichever of $x$ and $-x$ is nonnegative in the order $\leq\_R$. $\square$


So, as an ordered field, $R$ embeds in $\RR$. Let $\phi: R \to \RR$ be this embedding. Let $\RR^{alg}$ be the field of algebraic elements in $\RR$.


Since $R$ is algebraic over $\QQ$, we have $\phi(R) \subseteq \RR^{alg}$. Since no nontrivial finite extension of $R$ can be ordered (property 6 on Wikipedia's list of properties of real closed fields), we actually have $\phi(R) = \RR^{alg}$. We have $\bar{\QQ} = R(\sqrt{-1}) = \RR^{alg}(\sqrt{-1})$. So we can extend $\phi$ to an automorphism of $\bar{\QQ}$ by declaring it to fix $\sqrt{-1}$. Then $\phi$ conjugates $\sigma$ to complex conjugation.


