What is $\mathbb{C}^{\mathrm{Aut}(\mathbb{C}/\mathbb{Q})}$? Let $\mathrm{Aut}(\mathbb{C}/\mathbb{Q})$ be the field automorphisms of $\mathbb{C}$, and $\mathbb{C}^{\mathrm{Aut}(\mathbb{C}/\mathbb{Q})}$ the subfield of $\mathbb{C}$ fixed by this group. I supsect that it is equal to $\mathbb{Q}$ but I have difficulty proving it.
Here is what I do: let $x \notin \mathbb{Q}$. We have to find an automorphism not fixing $x$. It is easy to find an embeding $\mathbb{Q}(x) \rightarrow \mathbb{C}$ not sending $x$ on $x$, and use Zorn's lemma to extend it to a maximal subfield (of $\mathbb{C}$) $K \supset \mathbb{Q}(x)$. But it is not true that $K=\mathbb{C}$. So how to handle it?
 A: 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 $L=L_0$.)
A: 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.
A: I think that the article Chandru posted indicates that you can interchange any two irrational algebraic numbers with the same minimal polynomial or any two transcendental numbers with an automorphism of C. Thus the fixed field of Aut(C/Q) is exactly Q. The trick here is not to consider the poset of extension maps of Q(x) into C, but rather to start with an automorphism of the field and then look at the poset of automorphisms of some field extension. The punchline is towards the end of the article.
