I'm looking for a counterexample of the following (false) statement:
If $ K \subset M \subset L $ with $ [ L : K ] < \infty $ and $ |\text{Aut} (L / K)| = 1 $, then $ |\text{Aut} (M / K)| = 1 $.
Basically looking a finite algebraic field extension $ M / K $ with nontrivial $ K $-automorphisms which cannot be extended to $K$-automorphisms of $L$.
The Galois closure of $L$ has to be a bit complicated in order for counterexample to occur. Let $G$ be the subgroup of $S_6$ generated by $$\begin{pmatrix}1 & 2 &5 &4 &3 &6 \end{pmatrix} \qquad \begin{pmatrix}1 &5 \end{pmatrix} \begin{pmatrix}2 &6 \end{pmatrix}$$ it has order $36$. Stabilizer of $1$ is given by $$H = \{1, (35)(46), (246), (24)(35), (264), (26)(35)\}$$ Note that $G$ contains odd permutation, so $G\cap A_6 \lhd G$.
Now let $f \in K[x]$ be a degree $6$ separable polynomial such that its Galois group is $G$, $E$ be the splitting field of $f$. Let $L = E^H$, since $H$ is the stablizer of a point, we can assume $L = K(\alpha)$, where is $\alpha$ a root of $f$. Denote $M = E^{G\cap A_6}$, then $M\subset L$ and $[M:K]=|\text{Aut}(M/K)| = 2$. Next we claim $|\text{Aut}(L/K)|=1$, it suffices to show $L$ contains no other roots of $f$ except $\alpha$, this is true by looking at the elements of $H$: each indices $2,3,4,5,6$ are moved by $H$.
A concrete example: let $\alpha$ be a root of $x^6-x^3+2 = 0$, $$L = \mathbb{Q}(\alpha) \qquad M=\mathbb{Q}(\sqrt{-7}) \qquad K=\mathbb{Q}$$ The Galois group of this polynomial is easily seen to be $G$ above. Of course, a direct proof of $|\text{Aut}(L/K)| = 1$ is also not difficult by writing down roots explicitly.