local picture of ramification at a node Let $C/S$ be a semistable curve, and $G$ a finite group acting faithfully on $C/S$, whose order is invertible on $S$.
Let $P\in C$ be a geometric point with image $s\in S$, such that $P$ is a node in $C_s$. For short, let $A := \widehat{\mathcal{O}}_s$, and $R := \widehat{\mathcal{O}}_P$. Then, at $P$, we may fix "local coordinates" - ie, we fix an identification
$$R = A[[X,Y]]/(XY-a)$$
where $a\in m_A$ (the maximal ideal of $A$).
Let $x,y$ be the images of $X,Y$, then relative to these coordinates, any $A$-linear automorphism $\sigma$ which fixes the ideals $(x),(y)\subset R$ (which are called the "branches" of the node) is determined uniquely by a pair $(u,\gamma)\in R^\times\times A^\times$ with $\gamma a = a$, where
$$\sigma(x) = ux,\qquad \sigma(y) = \gamma u^{-1}y$$
Now suppose that the stabilizer $G_P\le G$ of $P$ is cyclic and fixes the branches, so any $g\in G_P$ is described by $(u,\gamma)$ as above.
On the bottom of page 44 of "Champs de Hurwitz", the author deduces that "this description, together with theoreme 90, allows us to assume that the action of $G_P$ is given by a pair of opposite characters $\chi,\chi^{-1}$ of order $e > 1$, via:
$$\sigma(x) = \chi(\sigma)x,\qquad \sigma(y) = \chi(\sigma)^{-1}y$$
"
It is not at all obvious to me why this should be true, or even how "theoreme 90" (presumably Hilbert's thm 90) would come in.
EDIT: I should add that the action of $G_P$ on the cotangent space of $C$ at $P$ has determinant 1.
 A: Let me give you a counter example, so you can see that your statement is missing a hypothesis or an assumption. The counter example is the semistable curve $V(X_1X_2)$ in the projective plane over a field $k$ with coordinates $X_0, X_1, X_2$. It has a unique node. Consider the automorphism $(X_0, X_1, X_2) \mapsto (\lambda X_0, \mu X_1, \nu X_2)$ for any $\lambda, \mu, \nu \in k^*$. Clearly this is not always of the form you desire.
So this answers your question completely: you are right that it is not obvious that it should be true and indeed it is not true in the setting you stated it in. (Please, please, please do not edit the question.) Go back and very carefully study all the assumptions made by the authors, read the references they give completely (I suggest starting with the Wewers reference they give because in my experience he writes very clearly, but I actually haven't read his thing myself), and then if you are still worried about this point, ask a local expert or email somebody you know who does this kind of algebraic geometry.
PS: I want to warn readers about the classification of automorphisms on a ring such as $A[[s, t]]/(st - a)$ say with $A$ complete local and $a \in \mathfrak m_A$. An automorphism not swapping the branches needn't be of the form stated in the question. In general one needs to do a change of coordinates before it has that form, if it is even possible, because it is not possible in general. This is already true in the simpler case where one has a smooth rational point over a field. For example $\mathbf{C}$-algebra $\mathbf{C}[[x]]$ has many automorphisms of order $2$ and many (infinite order) automorphisms which are not conjugate to scalar multiplication.
A: Okay! Finally I see how to prove this.
First note that any $r\in R$ has a unique expansion as:
$$r = r_0 + \sum_{i\ge 1}a_ix^i + \sum_{i\ge 1}b_iy^i\qquad r_0,a_i,b_i\in A$$
Since $R$ is a local ring (with maximal ideal $m_R := (x,y,m_A)$), the quotient $\varphi : R\rightarrow R/m_R \cong k$ is a local ring homomorphism, and hence $r\in R$ is a unit if and only if $\varphi(r)\in k$ is a unit - ie, if and only if $r_0\in A^\times$.
Note that $k = R/m_R = A/m_A$ is algebraically closed, since we are looking at geometric points.
Under our presentation of $R$, we find that for any $\sigma\in G_P$, there is a unique unit $u_\sigma\in R^\times$ and $\gamma_\sigma\in A^\times$ such that $\sigma(x) = u_\sigma x$ and $\sigma(y) = u_\sigma^{-1}\gamma_\sigma y$. Further, there is a unique $\chi(\sigma)\in k^\times$ such that we may write the action as
$$\begin{array}{ll} x\mapsto \chi(\sigma) u_\sigma x \\ 
y \mapsto \chi(\sigma)^{-1} u_\sigma^{-1}\gamma_\sigma y
\end{array}$$
where now $u_\sigma\equiv 1\mod m_R$. Since $\sigma$ is $A$-linear, $\chi(\sigma)$ defines a character $\chi : G_P\rightarrow k^\times$.
Now the map $\sigma\mapsto u_\sigma$ defines a 1-cocycle in $Z^1(G_P,R^\times)$. Ie, it satisfies:
$$u_{\sigma\tau} = u_\sigma\cdot\sigma(u_\tau)$$
Let $g$ be a generator of $G_P$, so that $G_P = \{1,g,g^2,\ldots,g^{e-1}\}$. For any $\sigma\in G_P$, consider the function $f_{u_\sigma} : R\rightarrow R$ given by
$$f_{u}(r) = \sum_{\sigma\in G_P}u_\sigma\sigma(r) = \sum_{i=0}^{e-1}u_{g^i}g^i(r) = r + u_gg(r) + g(u_g)u_gg(r) + \cdots + \left(g^{e-2}(u_g)\cdots g(u_g)u_gg(r)\right)$$
The critical observation is that here, each term in the above sum fixes the residue class of $r$ mod $m_R$ (ie, fixes $r_0$ in it the $x,y$-expansion of $r$), and hence $f_{u}(r) \equiv er_0\mod m_R$. Since $e$ is invertible in $A$ by hypothesis, $r_0\in A^\times\iff er_0\in A^\times$, so $f_{u}$ sends units to units. Then, the usual computation in the proof of Hilbert's theorem 90 shows:
Let $b\in R^\times$, then for all $\tau\in G_P$,
$$\tau(f_u(b)) = \sum_{\sigma\in G_P}\tau(u_\sigma)\tau(\sigma(b)) = \sum_{\sigma\in G_P}u_\tau^{-1}u_{\tau\sigma}\tau(\sigma(b)) = u_\tau^{-1}\sum_{\sigma\in G_P}u_{\tau\sigma}(\tau\sigma)(b) = u_{\tau}^{-1}f_u(b)$$
Thus, $u_\tau = \frac{f_u(b)}{\tau(f_u(b))} = \frac{\tau(f_u(b)^{-1})}{f_u(b)^{-1}}$. Let $\beta_u := f_u(b)$, then
$$\sigma(\beta_u x) = \frac{\sigma(\beta_u)}{\beta_u}\chi(\sigma)u_\sigma \beta_u x = \frac{\sigma(\beta_u)}{\beta_u}\chi(\sigma)u_\sigma \beta_u x = \frac{\sigma(\beta_u)}{\beta_u}\chi(\sigma)\frac{\beta_u}{\sigma(\beta_u)} \beta_u x = \chi(\sigma)\beta_u x$$
Thus, replacing $x$ by $\beta_u x$ will get us the desired statement for the action of $G_P$ on $x$.
For the action of $G_P$ on $y$, observe that on the geometric fiber above $s$, the cotangent space at $P$ (ie, the cotangent space of $R/m_AR$) is generated by $x,y$, upon which $\sigma$ acts via the matrix
$$\begin{bmatrix}
\chi(\sigma) & 0 \\
0 & \chi(\sigma)^{-1}\gamma_\sigma
\end{bmatrix}\mod m_AR$$
The requirement that the action have determinant one then implies that $\gamma_\sigma = 1\mod m_A$, and hence viewed in $R$, $\gamma_\sigma = 1\mod m_R$. This implies that the 1-cocycle $\sigma\mapsto u_\sigma^{-1}\gamma_\sigma$ also has the property that the function $f_{u^{-1}\gamma} : R\rightarrow R$ given by
$$f_{u^{-1}\gamma} := \sum_{\sigma\in G_P}u_\sigma^{-1}\gamma_\sigma\sigma $$
also takes units to units, and thus we may apply the same argument above to find a unit $\beta_{u^{-1}\gamma}\in R^\times$ such that
$$\sigma(\beta_{u^{-1}\gamma}y) = \chi(\sigma)\beta_{u^{-1}\gamma} y$$
which completes the proof.
