I just stumbled upon the definition of a map between to groups
$$\varphi:G \to H$$
being a 1-isomorphism. It means that it is bijective and satisfies
$$\varphi(g)^n=\varphi(g^n)$$
for all $g \in G$ and $n \in \mathbb Z$.
So my question is: What is the smallest example of two non-isomorphic 1-isomorphic groups $G$, $H$? Smallest here in terms of cardinality. For $|G|=|H|=16$ there are two such pairs. But maybe for even smaller cardinalities such pairs exist.
I don't want to know just the answer and proof by computer. I want to know if there is a simple argument why no smaller examples exist.
Best Answer
It is clear that if $\phi$ is a $1$-isomorphism, then the order of $\phi(g)$ must be equal to the order of $g$: for otherwise, if the order of $\phi(g)$ were smaller, say $k$, then $\phi(g^k) = \phi(g)^k = e$ with $g^k\neq e$ (and $\phi(e)=e$ because $\phi(e) = \phi(e^2) = \phi(e)^2$). (If the order of $g$ were smaller, then apply the argument to $\phi^{-1}$.) Thus, the multiset of orders of elements of $G$ must be equal to the multiset for $H$.
(These are also called the "order sequences" of the groups, the ordered list of element orders. I think it's true that two finite groups are $1$-isomorphic if and only if they have identical order sequences, but I don't have time to work it out right now)
This makes it fairly straighforward to discuss the possibilities left.
Up to isomorphism, there is a unique group of order $p$ and of order $pq$ with $p\lt q$ both prime and $p\nmid q-1$; so none of those orders work. That discards $|G|=2$, $3$, $5$, $7$, $11$, $13$, and $15$.
For $p$ a prime, groups of order $p^2$ are abelian; the two abelian groups of order $p^2$ are $C_p\times C_p$ and $C_{p^2}$, and so you cannot establish the desired bijection unless the groups are isomorphic. That gets rid of $|G|=4$ and $9$.
For groups of order $pq$ with $p\mid q-1$, the (unique) abelian group is cyclic of order $pq$, and the nonabelian group has no elements of order $pq$, which gets rid of $|G|=6$, $10$, and $14$.
For groups of order $8$, the nonabelian groups $D_4$ and $Q_8$ have different number of elements of order $2$ (the first has four, the latter has one). The abelian groups are $C_8$, $C_2\times C_4$, and $C_2\times C_2\times C_2$. The first is the only one with elements of order $8$, so it cannot be part of such a pair; the second has exactly three elements of order $2$ (namely $(x,y^2)$, $(x,e)$, and $(e,y^2)$), so it cannot be part of a pair with $D_4$ or with $Q_8$; and the last has no elements of order $4$, so it cannot be part of a pair with the nonabelian ones or abelian ones, either.
That leaves only groups of order $12$. We can ignore $C_{12}$; the only remaining abelian one is $C_2\times C_6$, which has six elements of order $6$, no elements of order $4$, two elements of order $3$, and three elements of order $2$. The nonabelian groups are $D_{6}$ (which has two elements of order $6$ and six elements of order $2$, so it cannot be paired with the abelian groups), the dicyclic group $\mathrm{Dic}_{12}=\langle x,a\mid a^6=e, x^2=a^3, xax^{-1}=a^{-1}\rangle$, and $A_4$ (which has no elements of order $6$, hence cannot be paired with any other group of order $12$). The dicyclic group has elements of order $4$ (the Sylow $2$-subgroups are cyclic), so it cannot be paired with the dihedral group; and it cannot be paired with $C_2\times C_6$ because the latter lacks elements of order $4$.
Thus, there area also no examples in order $12$, and the smallest are in order $16$.
Relevant to this question are this previous question and this previous question