Let $G$ be the group of affine linear maps over the Galois field $k=GF(16)$
of order $16$. The elements of $G$ are maps from $k$ to itself of the form
$x\mapsto ax+b$ where $a\in k^*$ and $b\in G$. Those with $a=1$ form
a normal elementary abelian subgroup~$H$. All nontrivial elements of $H$
are conjugate. Then $H$ contains lots of subgroups of order $4$, thirty-five
in all, each consisting of the identity and three elements of this conjugacy class
of involutions. But these are not all conjugate under $G$; it is clear that such a
subgroup has at most fifteen conjugates.