The eigenstates of a time-independent Hamiltonian
$$ H |\psi_j \rangle = E_j|\psi_j\rangle$$
have the usual rotating-phase time dependence in the Schrödinger picture:
$$ |\psi_j(t)\rangle = |\psi_j(t_0)\rangle \cdot \exp(E_j(t-t_0)/i\hbar) $$
However, your formula indicates that $H_0$ and $V$ are time-dependent. So if the state vector is an eigenvector of $H_0(t)$ at one moment $t$, it will almost certainly not be an eigenstate of $H_0(t')$ at another moment $t'$ – because the operator is probably changing in a way that doesn't preserve the eigenvectors' being eigenvectors. For that reason, the simple "changing phase" Ansatz isn't a solution to the Schrödinger's equation.
For a generic Hamiltonian $H_0(t)$, there can't exist any state vector $|\psi(t)\rangle$ that simultaneously solves Schrödinger's equation (with the Hamilonian $H$ or even with $H_0(t)$); and that remains an eigenstate of $H_0(t)$ at each moment of time.
Because you say that $H$ is time-independent and $H_0,V$ seem to be time-dependent, it seems that all the formulae you are describing are in the Heisenberg picture, not Schrödinger's picture. In the Heisenberg picture, the state vector is independent of time (a constant function of $t$). In the Schrödinger's picture, the dependence of $H_0$ and $V$ on time would have to be "explicit" (the dynamical dependence is encoded in the evolution of the state vector in this picture) and in that case, it would be unreasonable for the explicitly time-dependent parts of $H_0,V$ to cancel in $H$.
In the Heisenberg picture, the dynamical time dependence is included in the evolution of the operators,
$$H_0(t) = \exp(-Ht/i\hbar) H_0(0) \exp(Ht/i\hbar) $$
and similarly for $V$ instead of $H_0$ (and for any operator).