There may be some technical issues with the question, but hopefully what I mean is clear...

Let $k$ be a number field (or maybe any finitely generated field over $\mathbb{Q}$ of characteristic 0)

Let $k(\!(t)\!)$ be the field of Laurent series in $t$ with coefficients in $k$, and let $\Omega$ denote an algebraic closure of $k(\!(t)\!)$, and let $\overline{k}$ denote the algebraic closure of $k$ inside $\Omega$.

There is a natural map $$\rho : \mathrm{Gal}(\Omega/k(\!(t)\!))\rightarrow \mathrm{Gal}(\overline{k}/k)$$ given by restriction.

If $\{t_n\}_{n\ge 1}\subset \Omega$ satisfies $t_1 = t$, $t_n^d = t_{n/d}$ for all $d\mid n$, then we say that it is a *compatible system of roots of $t$*.

Given a compatible system of roots $\{t_n\}$, and a filtration $\overline{k} = \bigcup L$ by finite extensions of $k$, the tensor product $\left(\varinjlim_L L(\!(t)\!)\right)\otimes_{k(\!(t)\!)} \left(\varinjlim_n k(\!(t)\!)(t_n)\right)$ is a field isomorphic to $\Omega$, and thus we obtain an section of the map $\rho$ by having $\sigma\in \mathrm{Gal}(\overline{k}/k)$ act on the tensor product in the obvious way in the first factor, and trivially on the second factor. In particular, this action stabilizes the compatible system of roots $\{t_n\}$.

**Does every section of $\rho$ stabilize some compatible system of roots $\{t_n\}$?**

Note that the kernel of $\rho$ is the subgroup $\mathrm{Gal}(\Omega/E)$ where $E = \varinjlim_L L(\!(t)\!)$ as before, and the Galois group is isomorphic to $\widehat{\mathbb{Z}}$. The group $\mathrm{Gal}(\overline{k}/k)$ acts on $\mathrm{Gal}(\Omega/E) \cong \widehat{\mathbb{Z}}$ via the cyclotomic character, and if $k$ is finitely generated over $\mathbb{Q}$ then the only element of $\widehat{\mathbb{Z}}$ fixed by all of $\mathrm{Gal}(\Omega/E)$ is the identity. This implies that if a section of $\rho$ comes from a compatible system, then it must be unique.