I have recently learned that if the degeneracy is not lifted in the first order in degenerate perturbation theory, one has to diagonalize a different matrix that plays a similar role to the first order perturbation matrix (see answers here) in order to obtain the second order energy corrections, but more importantly, the correct zeroth order eigenstates.
My question is basically, is there an easier way to determine the correct zeroth order states, without resolving to the systematic way described above?
In particular, I know that in the "regular" first order degenerate perturbation theory, if one can find a symmetry of the perturbing Hamiltonian (i.e an operator $\hat{A}$ such that $[\hat{V},\hat{A}]=0$ where $\hat{H} = \hat{H}_0+\hat{V}$) then the simultaneous eigenstates of both $\hat{H}_0$ and $\hat{A}$ are the correct zeroth order states.
Does this still holds when the degeneracy is not lifted in the first order? I was convinced this was the case after working out some examples where it does, but I am no longer sure it works in general.
For concreteness, consider $$ \hat{H} =\hat{H}_0+\hat{V}=\left(\begin{matrix}0&0&0\\0&0&0\\0&0&E\end{matrix}\right)+\left(\begin{array}{ccc} 0&0&V\\0&0&V\\V&V&0\end{array}\right), V\ll E$$ By exactly diagonalizing $\hat{H}$ and expanding to zeroth order in $V/E$ we see that the correct zeroth order eigenstates are $$\rvert +\rangle = \frac{1}{\sqrt{2}}(\rvert 1\rangle + \rvert 2 \rangle), \rvert -\rangle = \frac{1}{\sqrt{2}}(\rvert 1\rangle - \rvert 2 \rangle), \rvert 3 \rangle$$
The fact that $\hat{V}$ seems to be invariant under $\rvert 1 \rangle \leftrightarrow \rvert 2\rangle$ does suggests that these are the correct zeroth order eigenstates, and indeed the operator $$\hat{A} =\left(\begin{matrix} 0&1&0\\1&0&0\\0&0&1\end{matrix}\right)$$ commutes with $\hat{V}$.