This question is a follow-up to [Are there infinitely many L-rigs?](https://mathoverflow.net/questions/372349/are-there-infinitely-many-l-rigs) which is already pretty convoluted.


Define the $\varphi$-evaluation morphism at a complex number $s$ as $\epsilon\_{\varphi,s}:F\mapsto \varphi(F)(s)$ where $F$ is a map from $\mathbb{C}$ to itself. Consider a "general covariance condition" to naturally constrain the elements $\varphi$ of the automorphism group $G\_{\mathcal{L}}$ of some L-rig $\mathcal{L}$ by requiring this group to be invariant under the map $\iota$ that maps $\varphi$ to $\varphi^{-1}$ as well as under any element $g$ of $\operatorname{Aut}(G\_{\mathcal{L}})$, so we require the equality $\epsilon\_{\varphi,s}=\epsilon\_{g(\varphi),s}$ to hold for all $(g,\varphi,s)$. That way we'd get in particular $\varphi(F)(s)=\varphi^{-1}(F)(s)$, so that every automorphism of a given L-rig $\mathcal{L}$ would be of order at most $2$. Can we prove that $G\_{\mathcal{L}}$ is itself of order at most $2$?



Structure of the automorphism group of an L-rig
This an answer following an argument from Wojowu: as we require the equality $\epsilon\_{g(\varphi),s}=\epsilon\_{\varphi,s}$ to hold for all $(g,\varphi,s)$, and thus for all $s$, this means that $g(\varphi)=\varphi$ for all $\varphi$, so that $g$ is the identity. So the automorphism group of $G\_{\mathcal{L}}$ is trivial and $G\_{\mathcal{L}}$ is of order at most $2$.


