This question is a follow-up to 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 $F$ if 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 $\vert G_{\mathcal{L}}\vert$ is itself of order at most $2$?