The current question is motivated by the possibility of existence of an analogy between the absolute Galois group of the rationals and the dihedral group of order $2n$ seen as the isometry group of a convex regular $n$-gon. Indeed, a theorem of Artin says that the only elements of finite order of the absolute Galois group of the rationals, denoted by $G_{\mathbb{Q}}$, are the identity and the complex conjugation.
So can one write $G_{\mathbb{Q}}\cong H\rtimes\operatorname{Gal}(\mathbb{C}/\mathbb{R})$ for some normal subgroup $H$ of $G_{\mathbb{Q}}$?
By "the" complex conjugation I think you mean fixing an embedding $\overline{\mathbb{Q}}\hookrightarrow\mathbb{C}$.
Otherwise, it doesn't make sense to view $\operatorname{Gal}(\mathbb{C}/\mathbb{R})$ as a subgroup of $G_\mathbb{Q}$.
Take any imaginary quadratic extension $E/\mathbb{Q}$ (e.g. $E = \mathbb{Q}[i]$) and let $H$ be the subgroup $G_E$ of $G_\mathbb{Q}$.
It is then a normal subgroup and we have the short exact sequence $$1\rightarrow G_E\rightarrow G_\mathbb{Q} \rightarrow \operatorname{Gal}(\mathbb{C}/\mathbb{R}) \rightarrow 1,$$ because the complex conjugation induces an isomorphism $\operatorname{Gal}(\mathbb{C}/\mathbb{R})\simeq \operatorname{Gal}(E/\mathbb{Q})$.
This sequence splits via the embedding of $\overline{\mathbb{Q}}$ into $\mathbb{C}$.