Is a regular Galois extension of constant Galois extension Galois? Let $V, W$ be two (smooth, irreducible) varieties defined over $\mathbb{Q}$ and let $\phi: V \rightarrow W$ be a morphism defined over $\mathbb{Q}$. Suppose there is a number field $K$, which is Galois over $\mathbb{Q}$, such that $\phi$ is a regular Galois cover over $K$; i.e. $K(V)/\phi^*(K(W))$ is Galois with $\overline{K} \cap K(V)=K$.
Question: Is the extension $K(V)/\phi^*(\mathbb{Q}(W))$ Galois?
This feels like it should be true since it is built out of a constant Galois extension and a regular Galois extension which don't interact and examples suggest this to be true with the Galois group being a semidirect product. I couldn't find any results in the literature though about this mixed case so would be glad of references as well.
As a simple example, one could take $V=W=\mathbb{P}^1$ with $\phi: x \mapsto x^n$; this induces a regular Galois cover over the cyclotomic field $K=\mathbb{Q}(\zeta_n)$ with the Galois group of $K(V)/\phi^*(\mathbb{Q}(W)$ being isomorphic to $C_n \rtimes \operatorname{Aut}(C_n)$.
 A: $K(V)/\Bbb{Q}(V)$ and $K(V)/\phi^*(K(W)))$ both Galois implies that the finite extension $K(V)/(\Bbb{Q}(V)\cap \phi^*(K(W)))$ is Galois.

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*If also $\overline{K} \cap \Bbb{Q}(V)=\overline{K} \cap \phi^*(\Bbb{Q}(W))$ then $K(V)/\phi^*(\Bbb{Q}(W))=K(V)/(\Bbb{Q}(V)\cap \phi^*(K(W)))$ is Galois.


*But $\overline{K} \cap K(V)=K$ is not enough. Try with $K=\Bbb{Q}(\zeta_3),\phi^*(\Bbb{Q}(W))=\Bbb{Q}(x)$, $\Bbb{Q}(V)=K(V)=K(x)[y]/(y^2-x^3-x-\zeta_3)$.
For $K(V)/\phi^*(\Bbb{Q}(W))$ to be Galois we need to find an automorphism $\sigma$ of $K(V)$ fixing $x$ and such that $\sigma(\zeta_3)=\zeta_3^2$. It means that $\sigma(y)$ is a root of $T^2-x^3-x-\zeta_3^2$ ie. $\sigma(K(V))$ is the function field of an elliptic curve with a different $j$-invariant, so it can't be that $\sigma(K(V))=K(V)$.
$K(V)/\phi^*(K(W))=K(x)[y]/(y^2-x^3-x-\zeta_3)/K(x)$ is Galois because it is a quadratic extension, on the underlying smooth projective curves the covering $[x:y:1]\to [x:1]$ has a few ramified points with $x=0$, and removing them from both curves gives your Galois covering of smooth affine varieties.
The morphism $V\to W$ defined over $\Bbb{Q}$ is obtained from $V:y^2-x^3-x-a=a^2+a+1=yu-1=0\subset \Bbb{A^4}$ and sending $(x,y,a,u)\to x$
