Given a finite group $G$, let $\{(1,1),(m_1,n_1),\ldots,(m_r,n_r)\}$ be the list of pairs $(m,n)$ in which $m$ is the order of some element, and $n$ is the number of elements with this order. The order of $G$ is thus $1+n_1+\cdots+n_r$, and the pair $(1,1)$ accounts for the neutral element.

Let $G,G'$ be two finite groups, with the same list. Is it true or not (I bet *not*) that $G$ and $G'$ are isomorphic ? If not, please provide a counter-exemple.

**Edit**. Nick's answer gives the correct terminology, of *conformal groups*. Ben's answer speaks of the refined notion of *almost conjugate subgroups*. Is there any other related notion ?

Grassmann equivalence. I don't know the origins of either piece of terminology (conformality/ Grassmann equivalence), but in any case I really don't like either of them! I much preferorder portraitwhich a couple of people have used in their answers below. $\endgroup$ – Nick Gill Nov 18 '15 at 16:17