Another question about Higman's paper In page 26 of his paper:
http://plms.oxfordjournals.org/content/s3-10/1/24.full.pdf

Higman says the following:
If $ N_1 , N_2 $ are subgroups of $\Phi(H) $ that are in the same equivalence classe under the automorphism group of $H$ , then the number in such an equivalence class is at most equal to the order of the automorphism group of $H/\Phi(H) $ . 
He is saying that such an automorphism must induce the identity on $H/\Phi(H)$ , but I can't understand this part...
Can someone please explain to me what exactly does Higman say in this paragraph? How does he count the number of groups in each equivalence class?
Thanks in advance! 
 A: This is the theorem in which Higman proves a lower bound on the number of isomorphism classes of $p$-groups $H$ in which $\Phi(H)$ is central and elementary abelian.
We have a $p$-group $H$ in which $H/\Phi(H)$ is elementary abelian of order $p^r$ and $\Phi(H)$ is central in $H$ and elementary abelian of order $p^R$.
We look at subgroups $N$ of $\Phi(H)$ of order $p^{R-s}$, and we are considering how many different isomorphism classes of quotients $H/N$ we get. To do this he splits the subgroups $N$ into equivalence classes, where each such class contains groups that give isomorphic quotients $H/N$. The displayed formula is $a/b$, where $a$ is the total number of subgroups of $\Phi(H)$ of order $p^{R-s}$ (which is the same as the number of order $p^s$), and $b$ is the order of the automorphism group of $H/\Phi(H)$, which he has proved is an upper bound on the order of each equivalence class.
To prove this upper bound, he says  first that, if $H/N_1 \cong H/N_2$, then there is an automorphism of $H$ that maps $N_1$ to $N_2$. So that would give $|{\rm Aut}(H)|$ as an upper bound on the size of the equivalence classes. But, since any automorphism that induces the identity on $H/\Phi(H)$ must also induce the identity on $\Phi(H)$ and hence fix all of the subgroups $N$, we get the smaller upper bound $|{\rm Aut}(H/\Phi(H)|$ for the equivalence class sizes, and that is what he uses as $b$ in the formula.
