Conjecture on unipotent tensor squares for GL(n,q)

Let $n$ be a positive integer, let $\mu=(\mu_1,\mu_2,\dots)$ be a partition of $n$, and let $\tau$ be a partition of $n$. For triples of partitions define

$$U_{(\mu,\mu,\tau)}(q):=\left\langle\mathcal{U}^{\mu}\otimes\mathcal{U}^{\mu}\otimes\mathcal{U}^{\tau},1\right\rangle_{{\rm GL}_n(\mathbb{F}_q)},$$

where $\mathcal{U}^{\lambda}$ denotes the unipotent character of ${\rm GL}_n(\mathbb{F}_q)$ indexed by $\lambda$. The unipotent tensor-square conjecture. (i) $U_{(\mu,\mu,\tau)}(q)\neq 0$ for every partition $\tau$ of $n$ if and only if $\mu_1\leq\lceil n/2\rceil$. (ii) If $\mu_1>\lceil n/2\rceil$, then

$$U_{(\mu,\mu,(1^n))}(q)=0.$$

This gives a conjectural necessary and sufficient condition for a unipotent character's tensor square to contain every unipotent character. The claim is verified experimentally for $n\leq 8$; the paper proves the staircase case but leaves the general criterion open.