Question: How many partitions of $n$ distinct objects are there, given that all parts are equinumerous? (Let us not consider the empty partition and the identity partition.)

The cases when $n$ is the unit, or when $n$ is prime, are trivial. In these cases we have $0$ partition and $1$ partition respectively.

So treat this question as asking about the number of such partitions for composite $n.$

In particular, the case when $n = p^2$ with $p$ prime is very interesting. For example, when $n = 4,$ we have $3$ such partitions. What of when $n$ is $9,$ or $25,$ or $49,$ and so on?

Thank you.

Answer: 
Number of ways to partition an $n$-set into subsets of equal size is tabulated at <http://oeis.org/A038041> where there are formulas, programs, and links to the literature.
Well, for $p^2$ we can divide into $p^2$ classes of $1$ objects (one way) or $1$ class of $p^2$ objects (ditto) of of $p$ classes of $p$ objects.

How many ways to do the latter? Well there are ${p^2 \choose p}$ ways to choose the first class and ${p^2-p\choose p}$ ways to choose the second and so one but the order of the classes doesn't matter so there will be $\frac {\prod\_{k=0}^{p-1}{p^2-kp\choose p}}{p!}$.

Example there are $\frac {{9\choose 3}{6\choose 3}}{3!} = \frac {\frac {9\*8\*7}6\frac {6\*5\*4}6}6= 280$

In general to find the number of partitions for a number $n$ you must calculate how many ways there are do a partition for each particular factoring of $n=a\*b$.

So the answer would be $\sum\limits\_{n = a\*b}\frac {\prod\limits\_{k=0}^{b-1}{n-ka \choose a}}{a!}$