This is lemma 1.15. of Deligne, Mumford's paper.
Let $X$ be an irreducible stable curve over an algebraically closed field, $\phi$ an automorphism on $X$ which induces the identity on $\text{Pic}^0X$.
Then is $\phi$ the identity on $X$?


Let $X'$ be the normalization and $\phi'$ the action of $\phi$ on $X'$. (induced by the universal property of the normalization)
The author says that choosing a singular point of $X$ and an prdering of its inverse image in $X'$ defines a distinct morphism from $\mathbb{G}\_m$ to $\text{Pic}^0X$, and therefore the inverse image $S$ of the singular locus of $X$ is pointwise fixed by $\phi'$.
But I can't understand this at all.
(And I don't know what $\mathbb{G}\_m$ is.) 



The injectivity of $\text{Aut}(X) \to \text{Aut} (\text{Pic}^0X)$ for a stable curve
Let $s$ be a double point of $X$, and let $p,q$ be the two points of $\pi ^{-1}(s)$ (ordering chosen). Let $\pi :X\_s\rightarrow X$ be the partial normalization of $X$ at $s$. There is an exact sequence of sheaves
$$1\rightarrow \mathcal{O}\_{X}^\*\rightarrow \pi \_\*\mathcal{O}\_{X\_s}^\*\xrightarrow{\ \varphi \ } \kappa (s)^\*\rightarrow 1$$here $\kappa (s)^\*$ is the skyscrapper sheaf over $s$ with stalk $k^\*$, and $\varphi $ maps a function $f$ to $f(p)/f(q)$. The coboundary of the associated long exact sequence gives an injective homomorphism of algebraic groups $h\_s:\mathbb{G}\_m\rightarrow \operatorname{Pic}^{\mathrm{o}}(X) $. If $\operatorname{Sing}(X)=\Sigma $, the homomorphisms $h\_s \ (s \in \Sigma )$ define an injective homomorphism $\mathbb{G}\_m^{\Sigma }\rightarrow \operatorname{Pic}^{\mathrm{o}}(X) $; if $\phi$ induces the identity 
on $\operatorname{Pic}^{\mathrm{o}}(X) $, it must therefore preserve $\Sigma $ pointwise, and also $S=\pi ^{-1}(\Sigma )$ because exchanging $p$ and $q$ exchanges $h$ and $h^{-1}$.


