Return to Revisions

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 $2$.