Let $K$ be a field. Let $f \in K[x]$ be an irreducible separable polynomial and let $E$ be the splitting field for $f$ over $K$. Let $\alpha \in E$ be a root of $f$.
By the isomorphism extension theorem, we know that $\forall \beta \in E$ root of $f$, there is an automorphism $\psi \in \operatorname{Aut} (E/K)$ (i.e. fixing $K$) such that $\psi(\alpha)=\beta$.
What could we say on how many these automorphisms are for each choice of $\beta$? If I am not wrong, they are at maximum $[E : K[\alpha]]$. Could they be less? If so, does the number depend on $\beta$?
EDIT:
$E/K$ is a Galois extension because $E$ is the splitting field of a separable polynomial over $K$
The automorphism group acts on the set of roots. In your case, this action is transitive and you are asking how many group elements map from one to the other. This number is the same for all $\alpha, \beta$ (that is one property of a transitive action and you can show it from the orbit-stabilizer theorem), so it suffices to consider the case $\alpha = \beta$.
Then we are asking how many automorphisms leave $\alpha$ fixed, meaning that they leave all of $K[\alpha]$ fixed. The number of these is equal to $[E : K[\alpha]]$, as you already mentioned, as you have a Galois extension in your case.
As a small disclaimer: My time studying Galois theory is already quite long past, so please check that you actually do have a Galois extension here.