Let $ M $ be a class of L-functions such that whenever $ F $ and $ G $ belong to $ M $, then so do their product $ F.G $ and their tensor product $ F\otimes G $ defined by $ F\otimes G : s\mapsto\sum\_{n>0}\frac{a\_{n}(F)a\_{n}(G)}{n^s} $ for $\Re(s)>1 $ if $ F : s\mapsto\sum\_{n>0}\frac{a\_{n}(F)}{n^s} $ and $ G : s\mapsto\sum\_{n>0}\frac{a\_{n}(G)}{n^s} $ if $ \Re(s)>1 $ . Suppose also that the constant map $ s\mapsto 1 $ and the Riemann Zeta function $\zeta $ belong to $ M $ .
An automorphism of $ M $ is a bijection of $ M $ that sends a primitive element (i.e irreducible for the product) to a primitive element and that commutes to both the usual and the tensor product.


Let's define for an element $ F $ of $ M $ and a field automorphism of $ C $ denoted by $ \sigma $ the map $ \Psi\_{\sigma} : F\mapsto F\_{\sigma}=\sum\_{n>0}\frac{\sigma(a\_{n}(F))}{n^s} $ if $ \Re(s)>1 $ . $ \Psi\_{\sigma} $ is an automorphism of $ M $ .


Let's now define the 'Galois group' of $ F \in M\setminus\{1,\zeta\}$ as the group $ \operatorname{Gal}(F) $ , under composition, of field automorphisms $ \sigma $ of $ C $ such that $ F\_{\sigma}=F $ . If $ G\in M $ is such that there exists $ \sigma $ such that $ G=F\_{\sigma}\neq F $ and $\operatorname{Gal(F)} =\operatorname{Gal}(G) $ then I managed to prove that this group is abelian. Is it finite ? 



Galois group of an L-function
Let $M$ be the set of finite products of Dirichlet $L$-functions. These surely form a class of $L$-functions as in the question. Now take some prime $p$ congruent to 1 mod 4 and let $\chi$ be one of the two Dirichlet $L$-functions of conductor $p$ and order 4 (the other one will then be $\overline{\chi}$). Let $F$ be $L(\chi,s)$ and let $G$ be $L(\overline{\chi},s)$. Then $Gal(F)=Gal(G)$ is the automorphisms of the complex numbers which leave $i$ fixed. This group is certainly not finite (indeed it is uncountably infinite). If $\sigma$ is complex conjugation then $G=F\_\sigma\not=F$, so there is a counterexample.


