I have a square matrix $Q$. (I can additionally assume that it is invertible ${\rm det}\,Q \ne 0$, but I don't see how this helps.) I define yet another square matrix $Q_2 = Q^2$. This new matrix is happen to be Hermitian $Q_2^\dagger = Q_2$. Here is the question: how can I describe the set of such $Q$'s whose squares are Hermitian?
Obviously, if $Q$ is Hermitian, $Q^2$ is also Hermitian. Additionally, if $Q$ is anti-Hermitian (that is, $Q^\dagger = - Q$), $Q^2$ is Hermitian. The most broad set of matrices I am able to come up with is this: $Q = U^\dagger D U$, where $U$ is unitary, while $$D = {\rm diag}\, (d_1,..., d_n),$$ where any $d_\alpha$ ($\alpha = 1,..., n$) is either purely real, or purely imaginary. Does this exhaust all possible $Q$'s with the desired property, or other possibilities are possible?
In response to comments/replies, I want to make the following modification to the original problem.
Matrix $Q_2$ is Hermitian. I want to describe its square roots (that is, all matrices $Q$, such that $Q^2 = Q_2$).
My improved guess is the following. If $$Q_2 = U^\dagger D_2 U,$$ where $U$ is unitary, and $D_2 = {\rm diag}\,(d^{(2)}_1,..., d^{(2)}_n)$ is diagonal, then $$Q = U^\dagger X^{-1} D X U,$$ where $$D = {\rm diag}\,(\pm \sqrt{d^{(2)}_1}, ..., \pm \sqrt{d^{(2)}_n})$$ while $X$ is an invertible matrix that commutes with $D_2$: $XD_2 = D_2 X$. Does this description exhaust all possible $Q$'s?