Here is a counterexample of order $32$.
$G$ and $H$ will each have $3$ elements of order $2$ and $28$ elements of order $4$. In both cases all three elements of order $2$ will have square roots. That insures that $F_G=F_H$. But in $G$ one of them will have $4$ square roots while the others each have $12$, and in $H$ one of them will have $20$ square roots while the others each have $4$. That rules out a $1$-isomorphism.
Let $Q$ be a quaternion group of order $8$ and let $C\subset Q$ be a (cyclic) subgroup of order $4$. Inside $Q\times Q$ there are three subgroups of index $2$ that contain $C\times C$. Let $G$ be $Q\times C$ and let $H$ be the one that is neither $Q\times C$ nor $C\times Q$.
