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 ?



Finite groups with elements of the same order
There are easy examples that are $p$-groups. For instance, the mod 3 Heisenberg group is the nilpotent group with presentation 
$\left < a,b,c \;\bigg |\, [a,b] = c, [a,c] = [b,c] = a^3 = b^3 = c^3 = 1 \right >$ has order 27, and all but the trivial element of order 3. This has the same order portrait as $C\_3^3$ where $C\_3 = \mathbb Z / 3\mathbb Z$ is the cyclic group of order 3.


