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 $p^2$, $p^3$.



Number of  Subgroups of p-Groups
This is just a matter of searching for a counterexample!


You will find that in GAP or Magma, SmallGroup(81,7) has four subgroups of order 27, two of which are isomorphic, but the other two are not isomorphic to any others.


It is possible that the answer might be yes for abelian groups.


