The time-dependent Schrodinger equation is given as (with $\hbar=1$): $$i\dfrac{d}{dt}\psi(t)=H(t)\psi(t)\ ,$$ where $\psi$ is some normalized vector and $H(t)$ is a Hermitian matrix with time-dependent elements.
Let $\Psi(t)=V(t)\psi(t)$, where $V(t)$ is unitary. It can be shown that the time-dependent Schrodinger equation in terms of $\Psi$ can be written as: $$i\dfrac{d}{dt}\Psi(t)=\left[VHV^{-1}-iV\dot{\left(V^{-1}\right)}\right]\Psi(t)\ ,$$ where the overdot indicates element-wise time derivative. Is it possible to find a $V$ such that this new Hamiltonian $VHV^{-1}-iV\dot{\left(V^{-1}\right)}$ is real symmetric?
A simple solution can be found when $H$ is 2 x 2, by assuming that $V$ is diagonal. But, this method fails for higher dimensional cases. Can it be done under some special conditions (without assuming $H$ is constant)? Can it be done if $V$ is invertible, but not necessarily unitary?