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?

2$\begingroup$ If I'm not mistaken, all elements of order 2 are conjugate to complex conjugation. The fixed fields are naturally realclosed. $\endgroup$ – S. Carnahan♦ Jul 30 '13 at 13:01

$\begingroup$ @S.Carnahan Yes, that is correct. $\endgroup$ – Kevin Ventullo Jul 30 '13 at 13:14

3$\begingroup$ More generally, if $K$ is a number field then the set of conjugacy classes of elements of order 2 in ${\rm{Gal}}(\overline{K}/K)$ is in natural bijective correspondence with the set of real embeddings of $K$. $\endgroup$ – user36938 Jul 30 '13 at 13:45

$\begingroup$ @S.Carnahan I claim that your comments are false. See my answer below and see if you agree. $\endgroup$ – David E Speyer Jul 30 '13 at 18:43

1$\begingroup$ @David Speyer: You have misread the question and the comments: we're all talking about involutions in absolute Galois groups of number fields, not in the automorphism group of $\mathbf{C}$. I recommend thinking about my comment on the link with real embeddings (which is a wellknown fact in number theory) since it is very cool. $\endgroup$ – user36938 Jul 30 '13 at 19:02
$\def\QQ{\mathbb{Q}}\def\RR{\mathbb{R}}$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 ArtinSchrier theorem, $R$ is real closed. In particular, it comes with a natural order $\leq_R$ defined by $a \leq_R b$ if $\sqrt{ba} \in R$.
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 $t^n + a_{n1} t^{n1} \cdots + a_0$ be the minimal polynomial of $t$ over $\QQ$. But then $$a_{n1} t^{n1} \cdots + a_0 \leq_R t^{n1} \sum  a_i  <_R t^n,$$ 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.