Does $\text{Gal}(\bar K/K)$ act naturally on roots of $p$-adic power series $f(x)$ in $m_{\mathbb{C}_p}$?

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Let $K$ be a finite extension of the $p$-adic numebr field $\mathbb{Q}_p$ with algebraic closure $\bar K$.

Let us denote the integers by $O$ and maximal ideal by $m$.

Let $\mathbb{C}_p$ be the $p$-adic completion of $\bar K$ with open unit disk $m_{\mathbb{C}_p}$.

Let $f(x)$ be a $p$-adic power series over the integers $O$ with Weierstrass degree $d$ and consider the set $$S:=\{x \in m_{\mathbb{C}_p}~:~ f(x)=0 \}.$$

Does $\text{Gal}(\bar K/K)$ act naturally on the set $S$ ?

By $p$-adic Weierstrass Preparation Theorem, there exists a distinhuished polynomial $g(x)$ of degree $d$ and a unit power series $u(x)$ such that $$f=gu.$$ Now the polynomial $g(x)$ has total $d$ roots counting multiplicity in the open unit disk $m_{\mathbb{C}_p}$, and these exhaust all roots of the original series $f(x)$ that lies in $m_{\mathbb{C}_p}$.

This means that the elements of the set $S$ are algebraic, since they are roots of the polynomial $g(x)$.

Therefore, $\text{Gal}(\bar K/K)$ acts naturally on the set $S$,

i.e., $\sigma(\alpha) \in S$ for $\sigma \in \text{Gal}(\bar K/K)$ and for $\alpha \in S$.

Is my argument right ?