If $(A \overline{A} )^n = I$, then $A$ is diagonalisable?

Question

We know that $A^n = I$ implies that $A$ is a diagonalisable, as the minimum polynomial must have distinct roots.

I'm wondering if $(A \overline{A} )^m = A \overline{A} ... A \overline{A} = I$ implies that same?

If not then a quick counterexample would be appreciated!

Answer 1

Unless I made a stupid mistake, here is an example.

$$\begin{bmatrix} i & 1 \\ 0 & i \end{bmatrix}\begin{bmatrix} -i & 1 \\ 0 & -i \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

P.S. The answer is trivially yes for real matrices.

Answer 2

Counterxample: $$A=\pmatrix{1&i\\0&1}.$$