If $G$ is a finite $p$-group of order $p^n$, then it is well known that for ($1\leq m\leq n$), number of subgroups of order $p^m$ is $1$(mod $ p$).

* Question:* Is it true that number of subgroups of order $p^m$, which are isomorphic within themselves, is $0$(mod $ p$) or $1$(mod $p$).

It looks to be true for groups of order $p^2$, $p^3$.