Let $X$ be a connected reduced stable curve of genus $g$ over an algebraically closed field $k$ of characteristic not dividing $|G|$. Let $G$ be a finite group acting faithfully on $X$.


Let $A := Aut(X\rightarrow X/G)$ be the group of automorphisms $\alpha : X\rightarrow X$ which induce the identity on $X/G$ and commutes with the action of $G$. Clearly the center $Z(G)$ can be embedded in $A$.


A [paper](https://perso.univ-rennes1.fr/matthieu.romagny/articles/champs_de_Hurwitz.pdf) (second to last paragraph on p136, $\S$8.2) I'm reading implies that $A$ might be strictly larger than $Z(G)$. I'd like to understand how this might be possible, and if it is, if it is still possible under the assumptions:


1. $G$ acts freely on the smooth locus of $X$.
2. $G$ acts with determinant 1 on the cotangent spaces of the nodes if it fixes the branches, and with determinant -1 if it swaps the branches.



Can a "stable" $G$-torsor have automorphisms which do not lie in the center $Z(G)$?
Assume $k$ has characterstic not $2$. Take two curves $Y$ and $Z$ of genus $> 1$ over $k$ each with an automorphism of order $2$ having the same number of fixed points. Say the fixed points are $y\_1, \ldots, y\_n$ on $Y$ and $z\_1, \ldots, z\_n$ on $Z$. Let $X$ be obtained by glueing $y\_i$ to $z\_i$ in $Y \amalg Z$ and let $G = \mathbf{Z}/2\mathbf{Z}$ with $1$ acting via our order $2$ automorphism on both $Y$ and $Z$. Then $A$ will contain at least $4$ elements. For example, you can use the order $2$ automorphism of $Y$ and the identity on $Z$ and this will commute with $G$.


