$\DeclareMathOperator\Aut{Aut}\newcommand{\alg}{\mathrm{alg}}\newcommand{\an}{\mathrm{an}}$Edited after Noam Elkies' comment: From what I understand (very little actually), there exist elliptic curves defined over some number fields $\mathbb{K}$ Galois over $\mathbb{Q}$ which are isogenous to all their Galois conjugates and such elliptic curves are called $\mathbb{Q}$-curves.
Since two elliptic curves are isogenous if and only if they share the same L-function, can one define the automorphism group of the L-function corresponding to such an elliptic $\mathbb{Q}$-curve in such a way that this group is actually isomorphic to $\operatorname{Gal}(\mathbb{K}/\mathbb{Q})$?
EDIT November 25th 2013: it seems rather clear that the automorphism group of the isogeny class of the $\mathbb{Q}$-curve $E$ contains $\operatorname{Gal}(\mathbb{K}/\mathbb{Q})$ as a subgroup. As the L-function $F$ of $E$ determines its isogeny class entirely, the automorphism groups of $F$ and of the isogeny class of $E$ should be isomorphic. Therefore it remains to show that such a group doesn't contain any element not arising from $\operatorname{Gal}(\mathbb{K}/\mathbb{Q})$.
EDIT December 6th 2013: There may be a way to provide a definite answer to my question. Suppose $g_1, g_2, \cdots g_n$ are elements of the considered automorphism group not arising from $\operatorname{Gal}(\mathbb{K}/\mathbb{Q})$ and let $G$ be the group generated by $\operatorname{Gal}(\mathbb{K}/\mathbb{Q})\cup \{g_{1},\cdots g_{n}\}$. Then, provided $G$ is realizable over $\mathbb{Q}$, there should exist a number field $\mathbb{L}$ containing $\mathbb{K}$ as a proper subfield such that the considered $\mathbb{Q}$-curve is actually defined over $\mathbb{L}$.
Edit April 1rst 2015: I copy and paste here the content of a cross-post MSE question which, after getting one up vote, has received one down vote et two votes to close without any explanation even though I did my best to improve it. So here it is:
"If I'm not mistaken, the isogeny class of $E/K$ contains an abelian group $Z^{r_{\alg}}$ which is the torsion free part of the Mordell-Weil group of $E/K$. According to the BSD conjecture, the quantities $r_{\alg}$ and $r_{\an}:=\min\{k, L^{(k)}_{E/K}(1)\neq 0\}$ should be equal.
What I would like to know is whether the set of sub $Z$-modules (taken in the increasing order of rank) of $Z^{r_{\alg}}$ can be viewed as a Galois covering of $Z$, and equivalently if the set $\{L_{E/K}^{(k)}, L_{E/K}^{(k)}(1)=0\}$ (taken in the increasing order of derivation order) can be viewed as a Galois covering of $L_{E/K}$.
If so, are these two Galois coverings isomorphic with Galois group $\operatorname{Gal}(K/Q)$?
Of course, the objects considered in the present question are to be defined as sets endowed with a coherent topology to be viewed as genuine topological spaces. Perhaps the Zariski topology would do the job considering $\operatorname{spec} Z$ instead of $Z$, and the set of local factors of $L_{E/K}$ (i.e those that appear in the Euler product thereof) instead of $L_{E/K}$ itself.
The key idea is to somehow identify:
the group $\Aut_{B}(X)$ of automorphisms of the covering $X$ of the base $B$ preserving $B$ (where $B$ stands for the topological space corresponding to $L_{E/K}$ and $X$ to the $r_{\an}$ topological spaces corresponding to $L_{E/K}, L'_{E/K},...,L^{(r_{\an}-1)}$ in this order) with $\Aut(L_{E/K})$ which, as showed in the link above, is isomorphic to $\operatorname{Gal}(K/Q)$,
the group $\Aut_{B'}(Y)$ of automorphisms of the covering $Y$ of the base $B'$ preserving $B'$ (where $B'$ stands for the topological space corresponding to $Z$ and $Y$ to the $r_{\alg}$ topological spaces corresponding to $Z, Z^2,...,Z^{r_{\alg}}$ in this order) with the automorphism group of the isogeny class of $E/K$ which, as showed in the link above, is isomorphic to $\operatorname{Gal}(K/Q)$."
If some people here ever vote to close, please give a good reason to do so.
Thanks in advance.
Edit May 13th 2016: following http://www.math.wisc.edu/~ellenber/MCAV.pdf, it seems that the automorphism group of a $\mathbb{Q}$-curve is actually isomorphic to $\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$. I find that rather interesting since I've been suspecting for long that if one calls 'L-function' any element of the set $M$ generated by the Selberg class and automorphic L-functions closed under both the classical product and the Rankin-Selberg convolution, then the set of automorphisms thereof as the group of bijections from $M$ to itself preserving this double structure should be isomorphic to the absolute Galois group of the rational. Still, I never managed to prove this. Can anyone please help me try to clear this out?
Edit September 25th 2022: in Grothendieck's conception of Galois theory, the absolute Galois group of a number field is the automorphism group of a fiber functor, which may help reconcile the two points of view (automorphism group of the $\mathbb{Q}$-curve and of the L-function).
