Hello,


assuming Selberg's orthonormality conjecture, let's consider bijective maps $f$ from Selberg's class $\mathcal{S}$ to itself such as:


1) $f$ maps a primitive function of $\mathcal{S}$ to a primitive function of $\mathcal{S}$,


2) $f$ maps a function of degree $d$ to a function of degree $d$,


3) for every $F$ in $\mathcal{S}$, $n\_{f(F)}=n\_{F}$, where $n\_{F}$ is the integer involved in Selberg's conjecture $A$,


4) if $F=F\_{1}^{e\_{1}}.F\_{2}^{e\_{2}}....F\_{k}^{e\_{k}}$, then $f(F)=f(F\_{1})^{e\_{1}}.f(F\_{2})^{e\_{2}}....f(F\_{k})^{e\_{k}}$ (so that $f$ is "strongly multiplicative"),


5) if $f$ verifies the above conditions, then so does the inverse of $f$.


The set of all such maps makes a group for the composition law. I would like to know whether this group is isomorphic to the absolute Galois group of the field of the rational numbers or not.


Thanks in advance.



Automorphisms of the Selberg class
Most elements in $\mathcal{S}$ probably don't have algebraic Dirichlet coefficients (even after scaling), e.g. this is conjectured for the $L$-functions of non-CM Maass forms. So I don't see that the Galois group in question acts on $\mathcal{S}$. I think it is fair to conjecture that the only automorphism of $\mathcal{S}$ is the identity.


