Edited after Noam Elkies' comment: From what I understand (very few 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 $\operatorname{Gal}(\mathbb{K}/\mathbb{Q})$? If so, can one draw a parallel between this notion of automorphisms of such an L-function and the one I defined in Automorphisms of an L-function where an automorphism of an L-function $F$ as an element of the Selberg class $\mathcal{S}$ is roughly speaking a completely multiplicative bijective map from $\mathcal{S}$ to itself that preserves $F$ and every invariant of $F$ (such as the degree, the $H$-invariants, etc)?

Thanks in advance.

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 $\operatorname{Gal}(\mathbb{K}/\mathbb{Q})$.

Apart from that, I think requiring that all invariants be preserved adds too much "rigidity", and one can probably loosen these requirements. In fact, I aim at proving that the so-called "strong" automorphisms (i.e. those that preserve all the invariants) preserving a given element $F$ of the Selberg class are either the identity or the complex conjugation. Maybe one way to prove this would be to show that a strong automorphism different from the aforementioned maps doesn't preserve the functional equation of $F$, hence leading to a contradiction (since it's been proved that two elements of the Selberg class with the same invariants share the same functional equation). I'd be very grateful to anyone who would manage to establish such a fact rigorously.

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 $\mathbb{L}$. What do you think of this?

dosatisfy it are called "$\mathbb Q$-curves".) – Noam D. Elkies Nov 24 '13 at 16:39