$\zeta I$ for each $n$-th root of unity $\zeta$, such that $\bar H$ has order $n^2$. This means that $H$ is of “central type” with cyclic center. Such $H$ exist for all $n$. For instance, one can take $\bar H = A \times A$, where $A$ is an abelian group of order $n$. Some nonabelian examples of such $\bar H$ are given by Iwahori and Matsumoto [@iwahori-matsumoto Sec 5]. Iwahori and Matsumoto conjectured that groups of central type are necessarily solvable and this was proved using the classification of finite simple groups by Howlett and Isaacs [@howlett-isaacs].

Since $|H| = n^3$ and $|Z(H)| = n$, the identity $\sum |\operatorname{tr}h|^2 = |H|$ implies $\operatorname{tr}h = 0$ for each $h \in H \setminus Z(H)$, i.e., the set of $h \in H$ such that $\operatorname{tr}h = 0$ has cardinality $n^3-n = (1 - \frac 1{n^2}) |H|$.

Let $G = H \times \{ \pm 1 \}$ and consider the representations of $G$ given by $\rho = \tau \otimes 1$ and $\rho' = \tau \otimes
\text{sgn}$, where $\tau$ is the standard representation of $H$ and $\text{sgn}$ is the nontrivial character of $\{ \pm 1 \}$. Then $\operatorname{tr}\rho(g) = \operatorname{tr}\rho'(g) = 0$ for $2(n^3-n) = (1 