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: 
If $m$ divides $n$, then there are $\frac{n!}{(m!)^{n/m}(n/m)!}$ to partition $n$ objects into parts each having size $m$. Proof: such a partition can be formed by ordering the $n$ objects in a row ($n!$ ways) and letting the first $m$ objects be the first part of the partition, the next $m$ objects being the second part, and so on. To account for overcounting, you must divide by $m!$ for each part (since you don't care about the order within each part of the partition) and divide by $(n/m)!$ (since you don't care about the order of the parts of the partition).

For instance when $n=4$ and $m=2$, we have $\frac{4!}{(2!)^2 (4/2)!} = \frac{24}{8}=3$. More generally, when $n=p^2$ and $m=p$, we have
$$\frac{(p^2)!}{(p!)^{p+1}}.$$

Then you have to sum over all divisors $m$ of $n$ (and exclude the case $m=1$ and $m=n$ if you want to avoid those trivial partitions).
$$\left(\sum\_{m \mid n} \frac{n!}{(m!)^{n/m}(n/m)!}\right) - 2$$
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.