If $D$ is a $3\times 3$-matrix of order $6$, then the geometric multiplicity of eigenvalue $-1$ in $D^3$ is two Consider the subgroup $H \le \operatorname{GL}(3,3)$ generated by the two matrices
$$
 A = \begin{pmatrix} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \end{pmatrix}
 \quad\mbox{ and }\quad
 B = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}
$$
If $D \in H$ has order $6$, then I want to show that $D^3$ is diagonalisable as
$$
 \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}
$$
i.e. the eigenvalue $-1$ has geometric multiplicity $2$. I know that as $D^3$ is involutory that $\pm 1$ are the only eigenvalues, also see this post. Also of course $x^6 - 1 = (x^3 - 1)(x^3 + 1) = (x-1)(x^2+x+1)(x+1)(x^2-x+1)$, so this polynomial is divided by the minimal polynomial of $D^6$, but I do not see if this might be helpful?

This answer deals with the specified group $H$.

The matrices $A$ and $B$ are both monomial matrices, i.e. they have a single non-zero entry on each row and column. Furthermore, they both have determinant one. Because monomial matrices form a group, and $SL_3(\Bbb{F}_3)$ is a group, we see that these two properties are shared by all the matrices in $H$.
Let's say that a matrix $D\in H$ belongs to the permutation $\sigma\in S_3$, if the non-zero entry of column $j$ appears on row $\sigma(j)$. It is easy to see that if $D$ belongs to the permutation $\sigma$, then $D^k$ belongs to the permutation $\sigma^k$. Therefore $\ell=\operatorname{ord}(\sigma)$ is the lowest power of $D$ that is diagonal. Because the non-zero elements of $\Bbb{F}_3$ are $\pm1$, we see that the order of $D$ is then either $\ell$ or $2\ell$.
We were given that $D$ has order $6$. Because $\ell\in\{1,2,3\}$ we must have $\ell=3$. And because $D$ had order $6$, we know that $D^3$ is a diagonal matrix with entries $\pm1$, not all $=1$. Because $\det D^3=1$, we can deduce that the number of entries $=-1$ in $D^3$ is exactly two. Q.E.D.
