Embeddings of $\overline{\mathbf{Q}}$ into $\mathbf{C}$ Keenan Kidwell's answer to Place stabilizers for the absolute Galois Group 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?   
 A: $\def\QQ{\mathbb{Q}}\def\RR{\mathbb{R}}$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, $R$ is real closed. 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.
