Order-cum-power statistics-equivalent not implies 1-isomorphic

History
This question was answered on Math Overflow here.

Statement
It is possible to have two groups $$G$$ and $$H$$ that are order-cum-power statistics-equivalent (i.e., they have the same order-cum-power statistics) but are not 1-isomorphic.

Proof
We construct groups $$G$$ and $$H$$ that are both of order $$32 = 2^5$$:


 * 1) $$G$$ is defined as the direct product of Q8 and Z4, i.e., it is the external direct product of the quaternion group and cyclic group of order 4.
 * 2) $$H$$ is defined as SmallGroup(32,35). One way of describing this group is as a maximal subgroup of direct product of Q8 and Q8 that is not isomorphic to direct product of Q8 and Z4.

Both $$G$$ and $$H$$ are groups of order $$32$$. They both have 1 identity element, 3 non-identity elements of order 2 that are squares, and 28 elements of order 4 that are not proper powers. Thus, they have the same order-cum-power statistics.

However, $$G$$ is not 1-isomorphic to $$H$$. To see this, note that the three non-identity elements of $$G$$ have $$12$$, $$12$$, and $$4$$ square roots. On the other hand, the three non-identity elements of $$H$$ have $$20$$, $$4$$ and $$4$$ square roots. Since the number of square roots of an element must be preserved under a 1-isomorphism, $$G$$ and $$H$$ are 1-isomorphic.