Let $L/F$ be a finite Galois extension. When $K$ is a normal intermediate subfield, $\operatorname{Aut}(K/F)$ is the quotient of $\operatorname{Aut}(L/F)$ by $\operatorname{Aut}(L/K)$. At first I thought that when $K$ was not normal, $\operatorname{Aut}(K/F)$ is the quotient of $\operatorname{Aut}(L/F)$ by the normal closure of $\operatorname{Aut}(L/K)$. But actually, that quotient is $\operatorname{Aut}(K'/F)$, where $K'$ is the largest normal subfield of $L$ contained in $K$.
So for example, if $L$ is the splitting field of $x^4-2$ over $\mathbb{Q}$, the quotient of $\operatorname{Aut}(L/\mathbb{Q})$ by the normal closure of $\operatorname{Aut}(L/\mathbb{Q}(\sqrt[4]{2}))$ is $\operatorname{Aut}(\mathbb{Q}(\sqrt{2})/\mathbb{Q})$.
(Correct me if I'm wrong about any of this.)
Question: Can someone give me a good example of a case where $\operatorname{Aut}(K/F)$ is no quotient of $\operatorname{Aut}(L/F)$? The case above doesn't quite work, because $\operatorname{Aut}(\mathbb{Q}(\sqrt[4]{2}/\mathbb{Q}) \cong \mathbb{Z}/2$ is a quotient of $D_8 \cong \operatorname{Aut}(L/\mathbb{Q})$.
$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$$\newcommand{\Q}{\mathbb{Q}}$$\newcommand{\Aut}{\mathrm{Aut}}$Let $F = \Q$, and take $L$ to be the splitting field of a suitable polynomial $f \in \Q[x]$ of degree $5$, such that $\Aut(L/F) \cong S_{5}$. (For instance, $f = x^{5} − 6 x + 3$ will work.)
Let $K$ be the field generated by three of the roots of $f$, say $\alpha_{1}, \alpha_{2}, \alpha_{3}$. This has degree at most (actually equal to, I believe) $5 \cdot 4 \cdot 3 = 60$ over $F$, so $K \subsetneq L$, and $\Aut(K/F)$ has order at most $\Size{K : F} \le 60$.
Now $\Aut(K/F)$ has order divisible by $3$, as it will contain a $3$-cycle from $S_{5}$ that permutes the three roots $\alpha_{1}, \alpha_{2}, \alpha_{3}$. But $S_{5} \cong \Aut(L/F)$ has no proper quotient of order divisible by $3$.