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 is to an earlier version of the question, when the subgroup was not specified.

Collecting some of my comments to an answer. The claim is false as stated.
The order six matrix
$$
D=\pmatrix{-1&1&0\cr0&-1&0\cr0&0&-1\cr}
$$
raised to third power gives $-I$ and thus has $-1$ as an eigenvalue of multiplicity three.
Similarly another order six matrix
$$
D=\pmatrix{1&1&0\cr0&1&0\cr0&0&-1\cr}
$$
has as the cube the diagonal matrix with $1$ occuring twice and $-1$ once.
These examples show that


*

*it is not necessarily the case that both $+1$ and $-1$ occur as eigenvalues (one of the formulations of the question from the comments)

*$D^3$ may have $-1$ as an eigenvalue with multiplicity one only.



Given this I'm still wondering what the question really was?

Observe that any Jordan block of size two or three can be written in the form
$$
J=rI+N,
$$
where $N^3=0$. Because we are in characteristic three and $I$ and $N$ commute, the binomial formula gives
$$
J^3=r^3I+3r^2N+3rN^2+N^3=r^3I.
$$
This helped me in constructing those examples.
