Let $K \supset \mathbb{Q}_p$ be the $p$-adic field with ring of integers $\mathcal{O}_K$ and maximal ideal $m_K$ and let $\bar K$ be algebraic closure of $K$ with maximal ideal $\bar{m}_K$. Consider the absolute Galois group $G=\text{Gal}(\bar K/K)$.
When does the absolute Galois group $G$ stabilize or fix the maximal ideal $\bar{m}_K$ ?
I wonder how this is possible. But I can describe one possible situation where it may happen:
Suppose we have a $p$-adic power series $f(x) \in \mathcal{O}_K[[x]]$ without constant term. So we can consider iterates $f^{\circ n}(x)$ of $f(x)$ and its roots in $\bar{m}_K$. So consider, $$\Lambda_n=\{x \in \bar{m}_K~|~f^{\circ n}(x)=0\}.$$ Consider the field extension $L=\bigcup_nK(\Lambda_n)$. Thus $L$ is an algebraic extension.
Then $\text{Gal}(L/K)$ stabilizes the zero set $\Lambda=\bigcup_n \Lambda_n \subset \bar{m}_K$, and hence $\text{Gal}(\bar K/K)$, at least, stabilizes the infinite set $\Lambda \subset \bar{m}_K$.
Am I right ? Please comment here.
But when does $\text{Gal}(\bar K/K)$ fixes $\bar{m}_K$ ?
Will the assumption $\text{Gal}(L/K)$ being abelian help to answer my question ?
Thanks for any comments