The fixed field of Aut(C/Q) is Q

Let $F\subset \mathrm{\Omega }$ be a field extension with $\mathrm{\Omega }$ algebraic over $F$, let $F_0$ be an intermediate field, and let ${\varphi }_0:F_0\to {\mathrm{\Omega }}^{\mathrm{\prime }}$ be an embedding into an algebraically closed field ${\mathrm{\Omega }}^{\mathrm{\prime }}$. Order all extensions $\varphi :E\to {\mathrm{\Omega }}^{\mathrm{\prime }}$ with $F_0\subseteq E\subseteq \mathrm{\Omega }$ by extension. A chain has the union embedding, so Zorn's lemma from [F3] gives a maximal one. If $E\mathrm{\ne }\mathrm{\Omega }$, choose $\alpha \in \mathrm{\Omega }\setminus E$; it is algebraic over $E$. Apply $\varphi$ to its minimal polynomial and choose a root $\beta \in {\mathrm{\Omega }}^{\mathrm{\prime }}$. The transported minimal polynomial is irreducible over $\varphi (E)$ because $\varphi$ is an isomorphism onto its image, so sending $\alpha$ to $\beta$ extends $\varphi$ to an embedding of $E(\alpha )$, contradicting maximality. Hence $\varphi$ extends to $\mathrm{\Omega }$. If ${\varphi }_0$ maps $F$ onto $F$ and both source and target are algebraic closures of $F$, this extension is onto: its image is algebraically closed and contains $F$, so it contains every root of every polynomial over $F$, hence all of $\mathrm{\Omega }$.

Use [F2] to view $\bar{\mathbb{Q}}$ as a subfield of $\mathbb{C}$. By [F1], choose a transcendence basis $B$ of $\mathbb{C}$ over $\bar{\mathbb{Q}}$. Then $\mathbb{C}$ is algebraic over $\bar{\mathbb{Q}}(B)$ and, being algebraically closed, is an algebraic closure of that field.

If $z$ is algebraic over $\mathbb{Q}$ but $z\mathrm{\notin }\mathbb{Q}$, its irreducible minimal polynomial has degree at least two and has distinct roots because its derivative is nonzero and relatively prime to it. Choose another root $z^{\mathrm{\prime }}\in \bar{\mathbb{Q}}$. By step 1.1, the embedding $\mathbb{Q}(z)\to \bar{\mathbb{Q}}$ sending $z$ to $z^{\mathrm{\prime }}$ extends to an automorphism of $\bar{\mathbb{Q}}$ over $\mathbb{Q}$. Extend it coefficientwise to $\bar{\mathbb{Q}}(B)$ while fixing $B$, then apply step 1.1 to extend it to an automorphism of $\mathbb{C}$. This automorphism moves $z$.

If $z$ is transcendental over $\mathbb{Q}$, it is transcendental over the algebraic extension $\bar{\mathbb{Q}}$ by transitivity in [F2]. Extend $\{z\}$ to a transcendence basis $B$ over $\bar{\mathbb{Q}}$ using [F1]. The assignment fixing $\bar{\mathbb{Q}}$ and every element of $B\setminus \{z\}$ while sending $z$ to $z+1$ defines an automorphism of $\bar{\mathbb{Q}}(B)$; its inverse sends $z$ to $z-1$. By step 1.1 it extends to an automorphism of $\mathbb{C}$, and it moves $z$.

Every complex number outside $\mathbb{Q}$ is either algebraic nonrational or transcendental over $\mathbb{Q}$, and steps 2.1, 2.2 exhibit an automorphism fixing $\mathbb{Q}$ that moves it. Every element of $\mathbb{Q}$ is fixed by definition. Thus the common fixed field is exactly $\mathbb{Q}$.