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.
You can arrange the numbers from $1$ to $p^2$ in $(p^2)!$ ways. Thinking of each group of $p$ numbers as a bucket in a partition, rearranging the numbers in each bucket does not change the partition ($p$ buckets, each of which can be rearranged in $p!$ ways), nor does rearranging the buckets themselves ($p!$ arrangements). So there should be
$$\frac{(p^2)!}{(p!)^p\cdot p!} = \frac{(p^2)!}{(p!)^{p+1}}$$
such partitions excluding the two trivial partitions.