I'm trying to show that $A^* = A^{-3}$ iff $A^* = A$ and $A^2 = I$ given $A$ is invertible and $A\in M_n(\mathbb{C})$ but I'm having trouble with manipulating $A$ and just don't know what's legal or not. $A^*$ here is the adjoint of $A$. Any help would be appreciated!
I believe it's possible to do just by algebraic manipulation on $A$, but super lost as to how to achieve it.
Just to be super clear. $A^*$, the adjoint of $A$ is the transpose of the conjugate of $A$.
On
Since $A^\ast=A^{-3}$, we see that $A^\ast$ commutes with $A$. Hence $A^2=(A^\ast A)^{-1}$ is self-adjoint and positive definite. It follows that $A^2=(A^2)^\ast=(A^\ast)^2=(A^{-3})^2$, meaning that $A^8=I$ or $$ (A^2-I)(A^2+I)(A^4+I)=0.\tag{1} $$ Since $A^2$ is positive definite, so are $A^2+I$ and $A^4+I$. Therefore $(1)$ implies that $A^2-I=0$. But then we also have $AA^\ast-I=0$ because $A^\ast=A^{-3}=(A^2)^{-1}A^{-1}=A^{-1}$. Hence $A=A^\ast$.
If $A^* = A^{-3}$ then $A^*A = AA^*$ ($A$ is normal), so by the Spectral Theorem, $A$ is diagonalizable. Do a change of basis so that $A$ is diagonal. And now you can check that the only complex numbers that satisfy $\bar\lambda = \lambda^{-3}$ are $\lambda = \pm 1$.
If $A$ is diagonal and the diagonal entries are $\pm 1$ then $A^* = A$ and $A^2 = I$.