transitive action on finite abelian subgroups

Question

Let G be a group and K a finite subgroup of G. Let H be some subgroup of the normalizer of K in G, and assume the action of H on K by conjugation is transitive on elements of K of same order. Does H also acts transitively on isomorphic abelian subgroups of K?

Answer 1

Here is a counterexample where $H$ can be taken as the full automorphism group of $K$. Necessarily, $K$ is nonabelian. Namely, let $$ K = \left\{ \begin{pmatrix} a & x \\ 0 & 1 \end{pmatrix} \mid a, x \in \mathbb{F}_{16}, a^3 = 1 \right\}. $$ ($K$ is a semidirect product of $C_3$ and $(\mathbb{F}_{16},+)$, and a subgroup of the $G$ in Keith Kearnes's answer in a comment.)
Then $\operatorname{Aut}(K)$ acts transitively on elements of the same order (even some small subgroups do). But $K$ has two essentially different sorts of subgroups of order $4$ and type $C_2\times C_2$: Namely, $5$ of these subgroups are normal in $K$, and the other $30$ are not normal.