I define a 1-isomorphism between two groups as a bijection that restricts to an isomorphism on every cyclic subgroup on either side. There are plenty of examples of 1-isomorphisms that are not isomorphisms. For instance, the exponential map from the additive group of strictly upper triangular matrices to the multiplicative group of unipotent upper triangular matrices is a 1-isomorphism. Many generalizations of this, such as the Baer and Lazard correspondences, also involve 1-isomorphisms between a group and the additive group of a Lie algebra/Lie ring.


Consider the following function *F* associated to a finite group *G*. For divisors $d\_1$, $d\_2$ of *G*, define $F\_G(d\_1,d\_2)$ as the number of elements of *G* that have order equal to $d\_1$ and that can be expressed in the form $x^{d\_2}$ for some $x \in G$.


Question: Suppose *G* and *H* are finite groups of the same order such that $F\_G = F\_H$. Does there necessarily exist a 1-isomorphism between *G* and *H*?


Note that the converse is obviously true: if there exists a 1-isomorphism between *G* and *H*, then $F\_G = F\_H$.


Incidentally, *just* knowing the orders of elements does not determine the group up to 1-isomorphism. There are many counterexamples of order 16, with two non-abelian groups (one being the direct product of the quaternion group and the cyclic group of order two, and the other a semidirect product of cyclic groups of order four) having the same statistics on orders of elements as $\mathbb{Z}\_4 \times \mathbb{Z}\_4$, but neither being 1-isomorphic to it because they don't have the same number of squares.


Similarly, just knowing how many elements are there of the form $x^d$ for each divisor *d* of the order is not sufficient to determine the group up to 1-isomorphism. Again, there are counterexamples of order 16.



Order information enough to guarantee 1-isomorphism?
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 $C\times Q$.


