I think there are such examples for all odd primes $p$ and all $n \ge 3$.

There is a $p$-group $P$ of exponent $p$ of class $2$, with $\Phi(P)=Z(P)$ and $P/\Phi(P)$ and $\Phi(P)$ elementary abelian, with $|\phi(P)| = p^{n(n-1)/2}$, $|P/\Phi(P)|=p^n$, such that ${\rm Aut}(P)$ acts on $P/\Phi(P)$ as ${\rm GL}(n,p)$, where the induced action on $\Phi(P)$ is as the exterior square of the natural module for ${\rm GL}(n,p)$.

The group $P$ is defined by the presentation $\langle X \mid R \rangle$, where $$X=\{x_i:1 \le i \le n\} \cup \{y_{ij}^p: 1\le i<j\le n\} $$
and $$R=\{x_i^p:1 \le i \le n\}\cup \{y_{ij}^p: 1 \le i < j \le n\} \cup \{[x_i,x_j]y_{ij}^{-1}: 1 \le i < j \le n\} \cup C,$$
where $C$ consists of all commutators of all $y_{ij}$ with all other generators (to make the $y_{ij}$ central).

Now, by a well-known result of Zigmundy, there is a prime $q$ that divides $p^{n}-1$ but does not divide $p^r-1$ for any $r<n$, and ${\rm GL}(n,p)$ has a cyclic subgroup $Q$ of order $q$ that must act irreducibly on the natural module.

Now all nontrivial irreducible modules for $Q$ over ${\mathbb F}_p$ have dimension $n$, and in particular the exterior square $E$ of the natural module has a quotient module $E/K$ of dimension $n$. Let $R$ be subgroup of $\Phi(P)$ corresponding to $K$, and $G=P/R$. Then $|G|=p^{2n}$ with $|\Phi(G)|=p^n$, and since the cyclic group $Q$ acts irreducibly on both $\Phi(G)$ and $G/\Phi(G)$, the only characteristic subgroups of $G$ are $1$, $\Phi(G)$ and $G$.