Let $k$ be a number field and $C$ a smooth projective curve over $k$. Fix a prime $p$ and two closed points $P,Q$ of $C$ of degree $p$, i.e., the residue fields $k(P)$ and $k(Q)$ are degree $p$ extensions of $k$. Fix a separable closure $k_s$ of $k$.
Are the groups $\Gamma_P = \mathrm{Gal}(k_s/k(P))$ and $\Gamma_Q = \mathrm{Gal}(k_s/k(Q))$ isomorphic?
Let $\Gamma_k$ denote the absolute Galois group of $k$. Are $\Gamma_P$ and $\Gamma_Q$ normal subgroups of $\Gamma_k$? Even if so, we have $$\Gamma_k/\Gamma_P \cong H \cong \Gamma_k/\Gamma_Q,$$ for $H$ a group of order $p$, but I know in general we don't have $\Gamma_P \cong \Gamma_Q$. Hence I'm asking to see if there are exceptions when considering profinite groups instead.
$Gal(\overline{\Bbb{Q}}/\Bbb{Q}(i))$ is not isomorphic to $Gal(\overline{\Bbb{Q}}/\Bbb{Q}(\sqrt2))$ as abstract groups. The latter contains an element of order $2$ (the complex conjugaison) whereas the former doesn't: if $\sigma$ has order $2$ then let $E=\overline{\Bbb{Q}}^\sigma$ the subfield fixed by $\sigma$, it is such that $[\overline{\Bbb{Q}}:E]=2$, due to https://kconrad.math.uconn.edu/blurbs/galoistheory/artinschreier.pdf it implies that $\overline{\Bbb{Q}} = E(i)$, a contradiction as $i\in E$.
So take $k=\Bbb{Q},p=2,C=\Bbb{P}_k^1,P=(x^2+y^2),Q=(x^2-2y^2)$.