Is there a way to classify all finite groups $G$, such that $\pi(G) := \Pi_{H \triangleleft G} |H| = |G|^2$?
For abelian groups this problem is quite simple. All such abelian groups are exactly $C_{p^3}$ and $C_{pq}$ for any primes $p$ and $q$:
Suppose $P$ and $Q$ are abelian groups of coprime order. Then if $P \times Q$ satisfies that condition, then $\pi(P)|Q|^2 \leq \pi(P \times Q) = |P \times Q|^2 = |P|^2|Q|^2$ and thus $\pi(P) \leq |P|^2$.
So, to solve the problem we need to obtain the classification of all finite abelian $P$-groups, satisfying the condition $\pi(P) \leq |P|^2$.
Suppose, a finite abelian $p$-group $P$ is a direct product of $n$ cyclic groups. Then it contains a subgroup isomorphic to $C_p^n$. So $\pi(P) \geq \pi(C_p^n) = \Pi_{k = 0}^n (p^k)^{C_n^k} = p^{\Sigma_{k = 0}^n kC_n^k} = p^{n2^{n-1}} > |P|^2$ for any $n > 1$. So $P$ has to be cyclic.
For cyclic $p$-groups $P$ we can see $\pi({C_{p^n}}) = p^{\frac{n(n+1)}{2}}$. So, the group satisfies our condition iff $\frac{n(n+1)}{2} \leq 2n$ which is exactly, when $n \leq 3$.
Now one can also see that $\pi(C_{p^2q}) = p^6q^3 \geq |\pi(C_{p^2q})|^2$ and $\pi(C_{pqr}) = (pqr)^3 \geq |\pi(C_{pqr})|^2$ for any prime $p$, $q$ and $r$. So, only possible examples are of one of the following forms: $C_{p^3}$, $C_{p^2}$, $C_{p}$, $E$, $C_{pq}$. And one can via manual checking see, that only $C_{p^3}$ and $C_{pq}$ do actually satisfy that condition.
However, I do not know, what to do there in non-abelian case.