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This an answer following an argument from Wojowu: as we require the inversion to be an automorphism of $G_{\mathcal{L}}$, and the equality $\epsilon_{g(\varphi),s}=\epsilon_{\varphi,s}$ to hold for all $(g,\varphi,s)$, it means that $G_{\mathcal{L}}$ is abelian and consists of elements of order at most $2$, so that $G_{\mathcal{L}}=(\mathbb{Z}/2)^{n}$ for some $n$. But as the considered equality holds 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 $2$.