If $A^3 = I$, then I want to find the possible Jordan forms of the matrix. Since the minimal polynomial has degree at most three, each block is at most 3, and the eigenvalues are third roots of unity.
Is that the answer, or are there some other components I am missing?
On
If you want to work only with real matrices, then you can separate out the real and imaginary part to get $$ M = \begin{pmatrix}1 & 0 & 0\cr 0 & -\frac{1}{2} & -\frac{\sqrt{3}}{2}\cr 0 & \frac{\sqrt{3}}{2} & -\frac{1}{2}\end{pmatrix}$$
You can always do a similarity transforms to get other solutions!
You're a little off there. Note that $A^3 = I$, which means that $A^3 - I = 0$, which means that the minimal polynomial $q_A(t)$ of $A$ divides $$ t^3 - 1= (t-1)(t-\omega)(t-\omega^2) $$ where $\omega=e^{2\pi i/3}$ is a root of unity. Note that the maximal degree of any factor of the minimal polynomial is $1$, which means that the maximum length Jordan block for any given eigenvalue is $1$.