If $\det A^{2k+1}=-\det(A^{2k})=-1$ for all $k\in \Bbb N_0$ and $\mathrm{tr}(A)=0$ then $A\in \mathbf{O}(n)$?

Question

Is it true that If $$\det A^{2k+1}=-\det(A^{2k})=-1,\qquad \mathrm{tr}(A)=0$$ for all $k\in \Bbb N_0$ then $A\in \mathbf{O}(n)$? What about $n=2m$?

Answer 1

No. As pointed out by another user in the comment, your conditions amount to $\det(A)=-1$ and $\operatorname{tr}(A)=0$. It is easy to cook up some counterexamples, such as $A=\pmatrix{1&1\\ 0&-1}$ or $A=\pmatrix{C&I\\ 0&P}$ where $C=\pmatrix{0&1&0\\ 0&0&1\\ -1&0&0}$ and $P=\pmatrix{0&1&0\\ 0&0&1\\ 1&0&0}$.