Can one characterize the elements $\alpha$ of $L=\overline{\mathbb F\_q(T)}\subset\Omega$, a completion of $L$ for the $\left(\frac1T\right)$-adic valuation such that for every conjugate $\beta$ of $\alpha$, there exists a continuous $\mathbb F\_q\left(\left(\frac1T\right)\right)$-isomorphism of $\Omega$ with $\sigma(\alpha)=\beta$.



Algebraic function fields with continuous automorphisms
Let $\Omega'$ be an algebraic closure of $\mathbb F\_q (( \frac{1}{T}))$. Then we can embed $\Omega'$ into $\Omega$ because every extension of $\mathbb F\_q (( \frac{1}{T}))$ is defined over $\mathbb F\_q(T)$, and $\Omega'$ is dense in $\Omega$ because it contains $L$, so therefore $\Omega$ is the completion of $\Omega'$. Thus every continuous automorphism of $\Omega'$ extends to a continuous automorphism of its completion $\Omega$. 


Thus it's equivalent to ask that every Galois conjugate of $\alpha$ is conjugate, over $\mathbb F\_q (( \frac{1}{T}))$, in $\Omega'$, to $\alpha$. Since $\Omega'$ is an algebraic closure of $\mathbb F\_q (( \frac{1}{T}))$, it's equivalent to ask that the minimal polynomial of $\alpha$ over $\mathbb F\_q(T)$ remains irreducible over $\mathbb F\_q((1/T))$.


There are many criteria for when a polynomial over a local field is / isn't irreducible, for instance in terms of the Newton polygon, so I think this is the best way to express it.


