# Natural numbers n which satisfy gnu(n)=n?

Are there any natural numbers $n$ (other than 1) for which $gnu(n)=n$? We define $gnu(n)$ to be the number of isomorphism classes of groups of order $n$. This question popped into my head today, and I couldn't come up with a proof one way or another.

In the paper by Conway, et al. entitled "Counting Groups: Gnus, Moas, and other Exotica, Conjecture 10.1 implies that there should be no such natural number $n$ which satisfies $gnu(n)=n$.

Does anybody have an argument to show that there is no natural number $n$ (other than 1) for which $gnu(n)=n$? If it is really straightforward, just say so and I'll work it out for myself when I find time. Thanks!

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Just for the sake of completeness: $\mathrm{gnu}(n)$ is the number of (isomorphism classes of) finite groups of order $n$. –  Guillaume Brunerie Nov 18 '10 at 1:40
Oops! Thanks for catching that. –  Glen M Wilson Nov 18 '10 at 1:44
In the paper you linked to they call such an $n$ group-perfect; they also state that they don't know if there is any group-perfect number other than 1. So I'm tempted to say that this is an open problem, though I'd be happy to be corrected. –  Faisal Nov 18 '10 at 2:12
If a 2008 paper (with John Conway as an author) says that the problem is open, then I think we can safely say that it is open and not at all trivial. –  Andy Putman Nov 18 '10 at 2:33
OK, I guess I should probably delve a bit deeper into that paper. Sorry for not doing my homework well enough! –  Glen M Wilson Nov 18 '10 at 2:36