Suppose $\mathbb{F}/\mathbb{K}$ is a field extension, and $f(x) \in \mathbb{K[x]}$ is minimal polynomial of $u \in \mathbb{F}$. Let $G=Aut_{\mathbb{K}}\mathbb{F}$ be the group of field automorphisms from $\mathbb{F}$ onto $\mathbb{F}$ which fix $\mathbb{K}$ elementwise, and let $H=Aut_{\mathbb{K(u)}}\mathbb{F}$.
We know $$[G: H]\le \text{# of distinct roots of $f(x)$ in $\mathbb{F}$},$$ since there is an injection map from the set of all left cosets of $H$ in $G$ to the set of all distinct roots of $f(x)$ in $\mathbb{F}$, given by $ \Omega: \sigma H \mapsto \sigma(u)$.
I want to know when $$ [G: H]= \text{# of distinct roots of $f(x)$ in $\mathbb{F}.$} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$$
I know Equality (*) holds for some cases.
Case 1: $\mathbb{F}$ is a splitting field of $f$ over $\mathbb{K}$.
If $v$ is another root of $f$, then there is an isomorphism $\tau: \mathbb{K}(u) \cong \mathbb{K}(v)$ with $\tau(u)=v$ and $\tau|_{\mathbb{K}}=id$. In this case, since $\mathbb{F}$ is a spliting field of $f$ over $\mathbb{K}$, it is a spliting field of $f$ over $\mathbb{K}(u)$ and over $\mathbb{K}(v)$. Hence, $\tau$ extends to an automorphism $\sigma \in Aut_{\mathbb{K}}\mathbb{F}$. Therefore, every distinct root of $f$ is image of some coset of $H$ under $\Omega$.
Case 2: $\mathbb{F}$ is a splitting field of the set $S$ of polynomials over $\mathbb{K}$, and $f$ splits over $\mathbb{F}$.
This case is similar to the previous case. We can consider $\mathbb{F}$ to be a splitting field of $S \cup {f}$ over $\mathbb{K}$. Hence, $\tau$ extends to an automorphism $\sigma \in Aut_{\mathbb{K}}\mathbb{F}$, where $\tau: \mathbb{K}(u) \cong \mathbb{K}(v)$ with $\tau(u)=v$ and $\tau|_{\mathbb{K}}=id$.
Now, if $f$ splits over $\mathbb{F}$ but $\mathbb{F}$ is not a splitting field of some set of polynomials, then what can we say?
Even in the more general case that $f$ does not split over $\mathbb{F}$, what can we say?
Are there some conditions that (*) holds?
Another case when this happens is $F = K(u)$, no matter how many roots $f(x)$ has in $F$.
If $v$ is a root of $f(x)$ in $F$ then $K(v) = K(u) = F$ since $f(x)$ is irreducible over $F$ and there's a unique $K$-automorphism $\sigma \colon F \to F$ where $\sigma(u) = v$. So we get a mapping $G \to \{{\rm roots \ of} \ f \ {\rm in} \ F\}$ by $\sigma \mapsto \sigma(u)$, which is surjective. We have $\sigma(u) = \tau(u)$ if and only if $\sigma = \tau$ on $F$ since $F = K(u)$, so $G$ is in bijection with the set of roots of $f(x)$ in $F$. In this case your group $H$ is trivial.
Example. In the field $\mathbf Q(\sqrt[4]{2})$, $x^4 - 2$ has two roots and the group $G = {\rm Aut}(\mathbf Q(\sqrt[4]{2})/\mathbf Q)$ has order $2$.