Galois action on automorphisms of a curve

Question

Let $C$ be a smooth projective curve defined over a local field $K/\mathbb{Q}_p$. Denote by $\text{Aut}(C)$ the geometric automorphism group of $C$, which consist of isomorphisms of $C\times_{\text{Spec}K}\text{Spec}\overline{K}$ with itself.

In Stoll's paper on rational points on twisting families of curves, see: paper, in Proposition 4.1, Stoll shows that if $\Gamma\le \text{Aut}(C)$ is a finite $K$-defined subgroup of the automorphism group of $C$, with $v(\#\Gamma) = 0$, then $I_K$ acts trivially on $\Gamma$.

My issue is this: I don't understand the statement. If $\Gamma$ consists of $K$-defined automorphisms of $C$, doesn't it mean that each $\gamma\in\Gamma$ is an isomorphism: $$\gamma: C/K\longrightarrow C/K,$$ and therefore, the algebraic equations defining $\gamma$ all have coefficients in $K$, which imply that $G_K$ acts on them trivially?

Answer 1

$K$-defined would mean that the subgroup itself is defined over $K$, not the elements.

Formally, this means that there are equations over $K$ which are satisfied by the coefficients of polynomials defining the automorphisms in the subgroup but not by automorphisms outside the subgroup.

Practically, it just means the subgroup is stable under the action of the absolute Galois group of $K$.