For odd $p$ and each $n\ge 1$ there are exactly two extraspecial groups of order $p^{1+2n}$, distinguished by their exponent

At $n=1$ an extraspecial group of order $p^3$ is nonabelian, hence isomorphic to $H_p$ or to $M_p$; their exponents are $p$ and $p^2$, so there are exactly two and the exponent tells them apart.

Assume, for every $m$ with $1\le m<n$: both exponents $p$ and $p^2$ are realised by extraspecial groups of order $p^{1+2m}$; every such group has an admissible family with at most one modular member; its exponent is $p$ when that number is zero and $p^2$ when it is one; and two such groups with the same exponent are isomorphic.

If $\varphi :G\to G^{\mathrm{\prime }}$ and $\psi :H\to H^{\mathrm{\prime }}$ are isomorphisms carrying the identified central subgroups onto the identified central subgroups compatibly with the identifying isomorphisms, then $\varphi \times \psi$ carries $N$ onto $N^{\mathrm{\prime }}$ and induces an isomorphism $G{\circ }_{\alpha }H\to G^{\mathrm{\prime }}{\circ }_{{\alpha }^{\mathrm{\prime }}}{H}^{\mathrm{\prime }}$.

Every $g\in P$ has $g^p\in P^p\le \mathrm{\Phi }(P)=Z(P)$, so $g^{p^2}=(g^p{)}^p=e$ and the exponent of $P$ divides $p^2$.

Let $P$ be extraspecial of order $p^{1+2n}$ with $n\ge 2$ and take an admissible family $P_1,\dots ,P_n$; each member is nonabelian of order $p^3$, hence isomorphic to $H_p$ or to $M_p$.

If two members $P_i,P_j$ are isomorphic to $M_p$, then $R=⟨P_i,P_j⟩$ is an internal central product of them, so $R\cong M_p\circ M_p$ and $R$ is an internal central product of a subgroup isomorphic to $H_p$ and one isomorphic to $M_p$, both with centre $Z(P)$; replacing $P_i,P_j$ by those two leaves an admissible family with one fewer modular member. Repeating, $P$ has an admissible family with $k\in \{0,1\}$ modular members.

If $k=0$ every member has exponent $p$; since the members commute and generate $P$ and the $p$-th power map is a homomorphism, every element of $P$ is a product of elements of the members and has $p$-th power the identity, so $\exp (P)=p$. If $k=1$ the modular member contains an element of order $p^2$, so $\exp (P)$ is a multiple of $p^2$, and by step 1.4 it equals $p^2$. Thus the exponent determines $k$.

Since $n\ge 2$ and $k\le 1$, some member $P_1$ is isomorphic to $H_p$; peeling it leaves $C=⟨P_2,\dots ,P_n⟩$, extraspecial of order $p^{1+2(n-1)}$ with an admissible family of $n-1$ members of which $k$ are modular, and $P\cong P_1{\circ }_{\mathrm{id}}C$ along the identity of $Z(P)$.

If $P$ and $P^{\mathrm{\prime }}$ are extraspecial of order $p^{1+2n}$ with the same exponent, their normalised families have the same $k$ by step 3.1, so the peeled subgroups $C$ and $C^{\mathrm{\prime }}$ have the same exponent and are isomorphic by the induction hypothesis. Any such isomorphism restricts to an isomorphism $Z(C)\to Z(C^{\mathrm{\prime }})$. For the peeled Heisenberg factors, the maps $(a,b,c)\mapsto (ra,b,rc)$ with $r\in {\mathbb{F}}_p^{\times }$ preserve the multiplication of [F7] and induce every automorphism of their order-$p$ centres; choose one whose central restriction makes the two factor isomorphisms compatible. Step 1.3 then induces $P\cong P^{\mathrm{\prime }}$.

Both exponents are realised: if $C$ is extraspecial of order $p^{1+2(n-1)}$ then $H_p\circ C$ is extraspecial of order $p^{1+2n}$. When $C$ has exponent $p$, the commuting generating images of [L13] and [L4] show that the product has exponent $p$; when $C$ has exponent $p^2$, its injective canonical image from [L13] still contains an element of order $p^2$, while step 1.4 bounds the product exponent by $p^2$. Applying this to the two groups supplied by the induction hypothesis gives one group of each exponent. With step 4.1 this gives exactly two isomorphism classes at order $p^{1+2n}$ and completes the induction.