“good” automorphisms of Galois classes of L functions

Question

This question is a follow-up to Galois classes of L-functions. My goal here is to make things clearer.

Definition 1

Let $A$ be a subclass of the Selberg class containing $s\mapsto 1$, closed under products, and such that every element of $A$ can be factored in a unique fashion in a product of primitive elements of $\mathcal{S}$, these primitive elements belonging to $A$. Such a subclass of $\mathcal{S}$ will be called a Galois class of L-functions.

Definition 2

Let $A$ be a Galois class of L-functions. $T$ is an automorphism of $A$ iff the following properties simultaneously hold true:

1) $T$ is a bijective map from A to itself
2) $T$ maps a primitive element of $A$ to a primitive element of $A$
3) for all $F$ in $A$, the degree of $F$ and the degree of $T(F)$ are the same
4) for all $F$, $G$ in $A$, $T(F.G)=T(F).T(G)$

Let $M$ be the maximal Galois class of L functions, $F$ an element of $M$. Let's denote $G_{S}(F)$ the group of complex isometries preserving globally the multiset of non trivial zeros of $F$, $G_{F}$ the group of automorphisms of $M$ preserving $F$.

Hadamard's factorization theorem says that knowing $F$ is equivalent to knowing its multiset of non trivial zeros. Therefore, preserving this multiset is equivalent to preserving $F$.
My question is: does this mean that among the automorphisms of $M$ preserving $F$, there exist "good" automorphisms of $M$ preserving $F$ such that if $G_{F}^{good}$ is the group of such "good" automorphisms, then there exists a group homomorphism $\rho: G_{F}^{good}\to G_{S}(F)$ that is actually an isomorphism?

Thanks in advance.

EDIT September 18th 2013: maybe one could establish the existence of the desired isomorphism asking the question in the framework of category theory, for example showing that the two categories $\mathcal{C}$ and $\mathcal{D}$ consisting of elements of $M$ or their multisets of zeros with the desired morphisms between these objects are equivalent, hence showing that the corresponding groups of automorphisms are isomorphic. In other words, I expect the groups $Aut_{\mathcal{C}}(F)$ and $Aut_{\mathcal{D}}(Z(F))$ for $F$ an object of $\mathcal{C}$ and $Z(F)$ the corresponding object of $\mathcal{D}$ to be, respectively, equal to $G_{F}^{good}$ and $G_{S}(F)$. Is there any hope to prove this rigorously?

Answer 1

The size of the automorphism group of a nontrivial Selberg $L$-function is at most $2$.

The reason is the distribution of zeroes of elements of the Selberg class. Such functions have no zeroes for $Re s>1$. For $Re s<0$, they have only the trivial zeroes. These live in a horizontal strip on linear progressions. In the critical strip, we have the non-trivial zeroes. There are infinitely many of these, and they are denser than any finite set of critical progressions.

Thus we can "see" the location of the critical strip and the region of trivial zeroes by looking at the function, so any automorphism must fix these locations. This implies that the automorphism cannot be a rotation, because otherwise it would not fix the direction of the axis running through the trivial zeros. It cannot be a a translation, because otherwise it would not fix the intersection of the trivial zeroes axis with the critical strip. It cannot be a reflection through any line other than a horizontal line through the trivial zeroes, because again it would not fix the direction of the trivial zeroes.

Hence it must be of the form $z \to \bar{z}+it$ for $t$ real. I am not sure whether we can rule out the case $t neq 0$. We can if it has a pole, since the pole must be unique and at $s=1$, but do all Selberg L-functions have poles? It sounds plausible but in my brief research I haven't seen a proof.

I'm not sure whether this means that there is an automorphism of the Galois class of L-functions. I don't see any reason why one couldn't have a conjugate-asymmetric class containing a conjugate-symmetric function.