Does every section of the map $Gal(\overline{k((t))}/k((t)))\rightarrow Gal(\overline{k}/k)$ stabilize a compatible system of roots of $t$?

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 : Gal(\Omega/k((t)))\rightarrow 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 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 $ Gal(\Omega/E)$ where $ E = \varinjlim_L L((t))$ as before, and the Galois group is isomorphic to $ \widehat{\mathbb{Z}}$ . The group $ Gal(\overline{k}/k)$ acts on $ 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 $ Gal(\Omega/E)$ is the identity. This implies that if a section of $ \rho$ comes from a compatible system, then it must be unique.