Let $\displaystyle A, B$ be finite groups with $\displaystyle H \subset A \cap B$ be subgroup.

Suppose there exists a finite number of elements $\displaystyle h_{1}, h_{2}, \ldots, h_{r} \in H$ such that $\displaystyle a \sim_{A} h_{1} \sim_{B} h_{2} \sim_{A} \ldots \sim_{A(B)} h_{r}$ where $\displaystyle a \in A$.

I know that $\displaystyle a, h_{1}, h_{2}, \ldots, h_{r}$ all have same order.

But it doesn't mean $\displaystyle a, h_{r}$ in the same conjugacy class.

Is there anyway to obtain $\displaystyle a \sim_{A} h_{r}$ or $\displaystyle a, h_{r}$ in the same conjugacy class without adding abelian or self-conjugate?