Are there infinitely many L-rigs?

Question

$\DeclareMathOperator{\Q}{\mathbb{Q}}$Call "L-rig" any class $\mathcal{L}$ of L-functions of automorphic representations of $\operatorname{GL}_{n}(\mathbb{A}_{\Q})$ for some $n$ belonging to the Selberg class that be closed under both the usual product (which we'll denote by $\times$) and the Rankin-Selberg convolution (which we'll denote by $\otimes$), containing the respective neutral elements $s\mapsto 1$ and $\zeta$, and fulfilling the abstract algebraic properties making $(\mathcal{L},\times,\otimes,s\mapsto 1,\zeta)$ a rig (ring without negatives).

Does the main result in automorphy of $m$-fold tensor products of GL(2), Dieulefait 2020 imply the existence of infinitely many non trivial L-rigs?

Moreover, denoting by $\mathcal{M}$ the maximal L-rig under inclusion, can we see it as the analogue for L-rigs of the separable closure of a field? If yes, would it make $\operatorname{Aut}(\mathcal{M})$ isomorphic to some absolute Galois group like, say, $\operatorname{Gal}(\bar{\Q}/\Q)$?

Edit October 25th, 2020: there are at least 3 different L-rigs, namely the trivial one $\mathcal{L}_{0}$ generated by $s\mapsto 1$ and the Riemann Zeta function, $\mathcal{M}$ and its sub-L-rig $\mathcal{D}$ consisting of all self-dual L-functions. Assuming $\operatorname{Aut}(\mathcal{M})$ is isomorphic to some absolute Galois group and the analogue for L-rigs of the fundamental theorem of Galois theory, this absolute Galois group can't be finite (as all such Galois groups are of order at most $2$). It may then be possible to prove that $\operatorname{Aut}(\mathcal{M})$ is profinite.

Edit October 30th, 2020: perhaps a way to show we face a profinite group would be to prove that $\displaystyle{\mathcal{M}}$ is defined by a filtration $(\mathcal{L}_{i}):={(\mathcal{L}(F_{i}))}_{i\in I}$ so that $\mathcal{M}=\varinjlim_{i\in I}\mathcal{L}_{i}$ and $\displaystyle{\operatorname{Aut}(\mathcal{M})\cong\varprojlim_{i\in I}\operatorname{Gal}\left(\frac{\mathcal{L}_{i}}{\mathcal{L}_{0}}\right)}$, where $\mathcal{L}(F)$ is the L-rig generated by $F$, the sequence of intermediate L-rigs $\mathcal{L}_{i}$ being analogues of Galois extensions of $\mathcal{L}_{0}$ defined above.

More exactly the considered Galois group should be $\operatorname{Gal}(\mathcal{K}_{\mathcal{L_{i}}}/\mathcal{K}_{\mathcal{L}_{0}})$ with $\mathcal{K_{L}}$ the field generated by the L-ring $\mathcal{L}$, that we can call an "L-field". Proving $\mathcal{K}_{\mathcal{L}_{0}}\cong\mathbb{Q}$ may imply that $\operatorname {Aut}(\mathcal{M})\cong\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$.

Edit November 1st, 2020: following the answer by nguyen quang do in
https://math.stackexchange.com/questions/2782069/abstract-properties-of-the-absolute-galois-group-over-mathbbq and assuming $\operatorname{Aut}(\mathcal{M})$ has the structure of an absolute Galois group, then it is a profinite group.

Edit November 11th 2020: as $\mathcal{L}_{0}$ is the L-ring generated by the neutral elements, $\mathcal{K}_{\mathcal{L}_{0}}$ is the L-field generated by those neutral elements, and as such is isomorphic to $\mathbb{Q}$. Now, the maximality of $\mathcal{M}$ implies that if the extension of $\mathbb{Q}$ isomorphic to $\mathcal{K}_{\mathcal{M}}$ is algebraic, then it is "its" algebraic closure $\bar{\mathbb{Q}}$.

Edit December 12th 2020: define the "symmetry group" $\operatorname{Sym}(F_{\pi})$ of an element $F_{\pi}:s\mapsto L(\pi,s)=\prod_{v}L_{v}(\pi,s)$ of $\mathcal{M}$ as the stabilizer thereof under the action of $\operatorname{Aut}(\mathcal{M})$ on $\mathcal{M}$. Then any permutation $\sigma$ of the places $v$ leaves $F_{\pi}$ invariant, so that if it induces an automorphism of $\mathcal{M}$, that we'll denote by $\phi_{\sigma}$, the latter induces an isomorphism between $\mathbb{Q}_{v}$ and $\mathbb{Q}_{\sigma(v)}$ as a morphism between fields. But $v\neq v'\Longrightarrow\mathbb{Q}_{v}\not\cong\mathbb{Q}_{v'}$ and in particular, $\sigma$ induces an automorphism of $\mathbb{R}$, (when $v$ is the archimedean place), hence either the identity or the complex conjugation. Hence $\operatorname{Sym}(F_{\pi})\cong\operatorname{Gal}(\mathbb{C}/\mathbb{R})$ if $\pi$ is self-contragredient, and is trivial otherwise.

Answer 1

The Rankin-Selberg convolution of a quadratic Dirichlet L-function with itself is the Riemann zeta function. Therefore the rig generated by $\{1, \zeta(s), L(s, \chi_d)\}$ consists of all finite products (and powers) of $\zeta(s)$ and $L(s, \chi_d)$. In particular, there are infinitely many L-rigs.

If you start with $\{1, \zeta(s), L(s, \chi)\}$ where $\chi$ is a primitive Dirichlet character, then Rankin-Selberg convolution gives you $L(s, \chi^j)$ for any positive integer $j$. That L-rig is generated by a finite set, depending on the order of $\chi$. So you get infinitely many non-isomorphic L-rigs.

Note that if $\chi^j$ is not primitive, then $L(s, \chi^j)$ should be interpreted as the Dirichlet L-function of the inducing primitive character.