What is $\mathbb{C}^{Aut(\mathbb{C}/\mathbb{Q})}$?

In this work in progress I state the following Fundamental Theorem of Galois Theory:

Let $K/F$ be any field extension. The following are equivalent:
(i) For all subextensions $L$ of $K/F$, $K^{\operatorname{Aut}(K/L)} = L$.
(ii) At least one of the following holds:
(a) $K/F$ is algebraic, normal and separable (i.e., a Galois extension in the usual sense), or
(b) $K$ is algebraically closed of characteristic zero.

That's the good news. The bad news is that this "Theorem", which I wrote down several years ago, is actually only a conjecture (it has remained as an open question on MathOverflow for more than a year, which is some indication of its nontriviality, at least). Okay, but there is more good news: the implication (ii) $\implies$ (i) is proven, and is a fairly routine application of basic field theory. Your question is a special case of (ii) $\implies$ (i), so there you go.

Added: Mea culpa, the proof of (ii) $\implies$ (i) in the linked to notes is "$\ldots$". (When you can't prove the big theorem you announce on the first page, you lose some motivation to fill in the other details, it seems.) Instead, please see $\S 10.1$ of my field theory notes in which there is a complete proof of (even a result slightly more general than) (ii) $\implies$ (i). Really, I promise.

The trick is to pick a transcendence basis. For instance, suppose $x\in\mathbb{C}$ is transcendental, and pick a transcendence basis $B$ for $\mathbb{C}$ which contains $x$. Let $y\in B$ be some element different from $x$. Then there is an isomorphism $f:\mathbb{Q}(B)\to\mathbb{Q}(B)$ which swaps $x$ and $y$ and fixes every other element of $B$. The isomorphism $f$ then extends to an isomorphism $\mathbb{C}\to\mathbb{C}$, since $\mathbb{C}$ is an algebraic closure of $\mathbb{Q}(B)$.

The case that $x$ is algebraic but irrational is similar. In that case we can first take a finite Galois extension $K$ of $\mathbb{Q}$ containing $x$ and an automorphism $f:K\to K$ which does not fix $x$. Picking a transcendence basis $B$ for $\mathbb{C}$, we can then extend $f$ to an isomorphism $K(B)\to K(B)$ which fixes each element of $B$. Finally, we can extend to $\mathbb{C}\to\mathbb{C}$, since $\mathbb{C}$ is an algebraic closure of $K(B)$.

(Here we are using the fact that if $k$ and $\ell$ are fields, $f:k\to\ell$ is an isomorphism, and $K$ and $L$ are algebraic closures of $k$ and $\ell$ respectively, then $f$ extends to an isomorphism $K\to L$. Indeed, we can take a maximal extension $f_0:K_0\to L_0$ of $f$ to an isomorphism between subfields of $K$ and $L$. If $a\in K\setminus K_0$, then $a$ is algebraic over $K_0$, so since $L$ is algebraically closed we can find $b\in L$ such that $f_0$ extends to $K_0(a)\to L_0(b)$. So we must have $K=K_0$, and a similar argument shows $L=L_0$.)