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Let $K$ be a finite extension of $\mathbb{Q}_p$ with ring of integers $\mathcal{O}$, maximal ideal $\mathfrak{m}$ and uniformizer $\pi$. Let $\bar K$ be the algebraic closure of $K$ and $\bar{\mathfrak{m}}$ be the integral closure of $\mathfrak{m}$.

Let $f(x) \in x \mathcal{O}[[x]]$ be a noninvertible power series with finite Weirstrass-degree and let $f^{\circ n}(x)$ be the $n^\text{th}$ iterate of $f(x)$. We denote the set of roots of $f^{\circ n}(x)$ by $S_n=\{x \in \bar{\mathfrak{m}}~|~f^{\circ n}(x)=0 \}$. Also denote $S=\bigcup_n S_n$.

Since $f$ has finite Weierstrass degree, $\#S_n$ is finite. Assume that the field extension $K(S_n)$ is Galois such that \begin{align}\operatorname{Gal}(K(S_n)/K) \cong \left( \mathcal{O}/\pi^n \mathcal{O} \right)^{\times}.\tag{1}\label{1}\end{align} Taking the inverse limit of the relation \eqref{1}, we get \begin{align} \operatorname{Gal}(\bar K/K) \cong \mathcal{O}^{\times}.\tag{2}\label{2} \end{align} There is a continuous surjection $\operatorname{Gal}(\bar K/K) \twoheadrightarrow \operatorname{Gal}(K(S_n)/K)$ by $\sigma \mapsto \sigma\rvert_{K(S_n)}$.

My question—

Suppose $g(x) \in x\mathcal{O}[[x]]$ be another power series so that \begin{align}\sigma(\alpha)=g(\alpha) ~~\text{for all $\alpha \in S_n$ and $\sigma \in \text{Gal}(K(S_n)/K)$} \tag{3} \label{3} \end{align} How can I extend the relation \eqref{3} to $$\sigma(\alpha)=g(\alpha)~~ \text{for all $\alpha \in S$ and $ \sigma \in \text{Gal}(\bar K/K)$}$$

Let $K$ be a finite extension of $\mathbb{Q}_p$ with ring of integers $\mathcal{O}$, maximal ideal $\mathfrak{m}$ and uniformizer $\pi$. Let $\bar K$ be the algebraic closure of $K$ and $\bar{\mathfrak{m}}$ be the integral closure of $\mathfrak{m}$.

Let $f(x) \in x \mathcal{O}[[x]]$ be a noninvertible power series with finite Weirstrass-degree and let $f^{\circ n}(x)$ be the $n^\text{th}$ iterate of $f(x)$. We denote the set of roots of $f^{\circ n}(x)$ by $S_n=\{x \in \bar{\mathfrak{m}}~|~f^{\circ n}(x)=0 \}$. Also denote $S=\bigcup_n S_n$.

Since $f$ has finite Weierstrass degree, $\#S_n$ is finite. Assume that the field extension $K(S_n)$ is Galois such that \begin{align}\operatorname{Gal}(K(S_n)/K) \cong \left( \mathcal{O}/\pi^n \mathcal{O} \right)^{\times}.\tag{1}\label{1}\end{align} Taking the inverse limit of the relation \eqref{1}, we get \begin{align} \operatorname{Gal}(\bar K/K) \cong \mathcal{O}^{\times}.\tag{2}\label{2} \end{align} There is a continuous surjection $\operatorname{Gal}(\bar K/K) \twoheadrightarrow \operatorname{Gal}(K(S_n)/K)$ by $\sigma \mapsto \sigma\rvert_{K(S_n)}$.

My question—

Suppose $g(x) \in x\mathcal{O}[[x]]$ be another power series such that there exists a $\sigma \in \text{Gal}(K(S_n)/K)$ satisfying \begin{align}\sigma(\alpha)=g(\alpha) ~~\text{for all $\alpha \in S_n$} \tag{3} \label{3} \end{align} How can I extend the relation \eqref{3} to $$\sigma(\alpha)=g(\alpha)~~ \text{for all $\alpha \in S$ and $ \sigma \in \text{Gal}(\bar K/K)$}$$

Let $K$ be a finite extension of $\mathbb{Q}_p$ with ring of integers $\mathcal{O}$, maximal ideal $\mathfrak{m}$ and uniformizer $\pi$. Let $\bar K$ be the algebraic closure of $K$ and $\bar{\mathfrak{m}}$ be the integral closure of $\mathfrak{m}$.

Let $f(x) \in x \mathcal{O}[[x]]$ be a noninvertible power series with finite Weirstrass-degree and let $f^{\circ n}(x)$ be the $n^\text{th}$ iterate of $f(x)$. We denote the set of roots of $f^{\circ n}(x)$ by $S_n=\{x \in \bar{\mathfrak{m}}~|~f^{\circ n}(x)=0 \}$. Also denote $S=\bigcup_n S_n$.

Since $f$ has finite Weierstrass degree, $\#S_n$ is finite. Assume that the field extension $K(S_n)$ is Galois such that \begin{align}\operatorname{Gal}(K(S_n)/K) \cong \left( \mathcal{O}/\pi^n \mathcal{O} \right)^{\times}.\tag{1}\label{1}\end{align} Taking the inverse limit of the relation \eqref{1}, we get \begin{align} \operatorname{Gal}(\bar K/K) \cong \mathcal{O}^{\times}.\tag{2}\label{2} \end{align} There is a continuous surjection $\operatorname{Gal}(\bar K/K) \twoheadrightarrow \operatorname{Gal}(K(S_n)/K)$ by $\sigma \mapsto \sigma\rvert_{K(S_n)}$.

My question—

Is the absolute Galois group $\operatorname{Gal}(\bar K/K)$ isomorphic to $\operatorname{Gal}(K(S)/K)$?

Edit: Looking the nice comments by @Will Sawin, @LSpice and @Kconrad, I understand that the left side of \eqref{2} should not have been written and it is not true in general. That reduces the question to

When will we get $\text{Gal}(\bar K/K)=\text{Gal}(K(S)/K\left(=\text{Gal}(K(\bigcup_nS_n)/K)\right)$ ?

How to obtain $\bar K$ here ? Is it possible to obtain $\bar K$ by adjoining all roots of all iterates of $f(x) \in x \mathcal{O}[[x]]$ ?

Let $K$ be a finite extension of $\mathbb{Q}_p$ with ring of integers $\mathcal{O}$, maximal ideal $\mathfrak{m}$ and uniformizer $\pi$. Let $\bar K$ be the algebraic closure of $K$ and $\bar{\mathfrak{m}}$ be the integral closure of $\mathfrak{m}$.

Let $f(x) \in x \mathcal{O}[[x]]$ be a noninvertible power series with finite Weirstrass-degree and let $f^{\circ n}(x)$ be the $n^\text{th}$ iterate of $f(x)$. We denote the set of roots of $f^{\circ n}(x)$ by $S_n=\{x \in \bar{\mathfrak{m}}~|~f^{\circ n}(x)=0 \}$. Also denote $S=\bigcup_n S_n$.

Since $f$ has finite Weierstrass degree, $\#S_n$ is finite. Assume that the field extension $K(S_n)$ is Galois such that \begin{align}\operatorname{Gal}(K(S_n)/K) \cong \left( \mathcal{O}/\pi^n \mathcal{O} \right)^{\times}.\tag{1}\label{1}\end{align} Taking the inverse limit of the relation \eqref{1}, we get \begin{align} \operatorname{Gal}(\bar K/K) \cong \mathcal{O}^{\times}.\tag{2}\label{2} \end{align} There is a continuous surjection $\operatorname{Gal}(\bar K/K) \twoheadrightarrow \operatorname{Gal}(K(S_n)/K)$ by $\sigma \mapsto \sigma\rvert_{K(S_n)}$.

My question—

Suppose $g(x) \in x\mathcal{O}[[x]]$ be another power series so that \begin{align}\sigma(\alpha)=g(\alpha) ~~\text{for all $\alpha \in S_n$ and $\sigma \in \text{Gal}(K(S_n)/K)$} \tag{3} \label{3} \end{align} How can I extend the relation \eqref{3} to $$\sigma(\alpha)=g(\alpha)~~ \text{for all $\alpha \in S$ and $ \sigma \in \text{Gal}(\bar K/K)$}$$

Let $K$ be a finite extension of $\mathbb{Q}_p$ with ring of integers $\mathcal{O}$, maximal ideal $\mathfrak{m}$ and uniformizer $\pi$. Let $\bar K$ be the algebraic closure of $K$ and $\bar{\mathfrak{m}}$ be the integral closure of $\mathfrak{m}$.

Let $f(x) \in x \mathcal{O}[[x]]$ be a noninvertible power series with finite Weirstrass-degree and let $f^{\circ n}(x)$ be the $n^\text{th}$ iterate of $f(x)$. We denote the set of roots of $f^{\circ n}(x)$ by $S_n=\{x \in \bar{\mathfrak{m}}~|~f^{\circ n}(x)=0 \}$. Also denote $S=\bigcup_n S_n$.

Since $f$ has finite Weierstrass degree, $\#S_n$ is finite. Assume that the field extension $K(S_n)$ is Galois such that \begin{align}\operatorname{Gal}(K(S_n)/K) \cong \left( \mathcal{O}/\pi^n \mathcal{O} \right)^{\times}.\tag{1}\label{1}\end{align} Taking the inverse limit of the relation \eqref{1}, we get \begin{align} \operatorname{Gal}(\bar K/K) \cong \mathcal{O}^{\times}.\tag{2}\label{2} \end{align} There is a continuous surjection $\operatorname{Gal}(\bar K/K) \twoheadrightarrow \operatorname{Gal}(K(S_n)/K)$ by $\sigma \mapsto \sigma\rvert_{K(S_n)}$.

My question—

Is the absolute Galois group $\operatorname{Gal}(\bar K/K)$ isomorphic to $\operatorname{Gal}(K(S)/K)$?

Let $K$ be a finite extension of $\mathbb{Q}_p$ with ring of integers $\mathcal{O}$, maximal ideal $\mathfrak{m}$ and uniformizer $\pi$. Let $\bar K$ be the algebraic closure of $K$ and $\bar{\mathfrak{m}}$ be the integral closure of $\mathfrak{m}$.

Let $f(x) \in x \mathcal{O}[[x]]$ be a noninvertible power series with finite Weirstrass-degree and let $f^{\circ n}(x)$ be the $n^\text{th}$ iterate of $f(x)$. We denote the set of roots of $f^{\circ n}(x)$ by $S_n=\{x \in \bar{\mathfrak{m}}~|~f^{\circ n}(x)=0 \}$. Also denote $S=\bigcup_n S_n$.

Since $f$ has finite Weierstrass degree, $\#S_n$ is finite. Assume that the field extension $K(S_n)$ is Galois such that \begin{align}\operatorname{Gal}(K(S_n)/K) \cong \left( \mathcal{O}/\pi^n \mathcal{O} \right)^{\times}.\tag{1}\label{1}\end{align} Taking the inverse limit of the relation \eqref{1}, we get \begin{align} \operatorname{Gal}(\bar K/K) \cong \mathcal{O}^{\times}.\tag{2}\label{2} \end{align} There is a continuous surjection $\operatorname{Gal}(\bar K/K) \twoheadrightarrow \operatorname{Gal}(K(S_n)/K)$ by $\sigma \mapsto \sigma\rvert_{K(S_n)}$.

My question—

Is the absolute Galois group $\operatorname{Gal}(\bar K/K)$ isomorphic to $\operatorname{Gal}(K(S)/K)$?

Edit: Looking the nice comments by @Will Sawin, @LSpice and @Kconrad, I understand that the left side of \eqref{2} should not have been written and it is not true in general. That reduces the question to

When will we get $\text{Gal}(\bar K/K)=\text{Gal}(K(S)/K\left(=\text{Gal}(K(\bigcup_nS_n)/K)\right)$ ?

How to obtain $\bar K$ here ? Is it possible to obtain $\bar K$ by adjoining all roots of all iterates of $f(x) \in x \mathcal{O}[[x]]$ ?