So I am aware that not every permutation of roots in a splitting field necessarily induces an automorphism, even if the permutation occurs amongst roots sharing the same minimal polynomial
Take $\Bbb{Q}(\zeta_n)$ for example, the permutation of $\zeta_n$ completely determines where the rest of the roots of $x^n-1$ is sent to, so not every permutation induces an automorphisms.
However, in this following case, is it true?
If I have two field, L and K Galois over k, and $L \cap K = k$ and $K = k(\alpha_1,....\alpha_n)$ such that $\alpha_i$ are roots to a separable polynomial over $k$, so that $K$ is its splitting field, then is it possible to prove with brute force that for $L(\alpha_i...)$, every permutation that induces an automorphism on $K$ over k also induces an automorphism for $L(\alpha_i...)$ over $L$?
I accomplished this by considering $KL(\alpha_i) = L(\alpha_i)$ as the splitting field of $f(x)$ over $L$. and the restriction of $Gal(L(\alpha_i)/L)$ into $Gal(K/k)$.
But I really want to try to prove this without resorting to Galois theory, just by examining the permutation of roots and show that if it is allowed in $K$, it is allowed in $L(\alpha_i....)$.