Finding correct zeroth order eigenstates in degenerate perturbation theory when degeneracy is not lifted by first order correction 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 $\hat{V}$.
 A: $\hat A$ shares eigenstates with both $\hat H_0$ and $\hat H$, but they are not the same eigenstates! 
Specifically, e.g. this one, you may normalize $\hat H$ by E, and set $V/E \equiv g$, to see that the (tilded) eigenvectors of $\hat H$ are
$$
|\tilde{-}\rangle= |-\rangle, \qquad |\tilde{+}\rangle= N_+ \left (|+ \rangle +\frac{1-\sqrt{1+8g^2}}{2\sqrt{2}g}|3\rangle \right ), \\
|\tilde{3}\rangle=N_3\left (\frac{1+\sqrt{1+8g^2}}{2}|3\rangle+\sqrt{2} g |+\rangle \right ),
$$
all with different eigenvalues 0,$  \frac{1}{2}\left(1 - \sqrt{1+8g^2}\right) \approx-4g^2$, and $ \frac{1}{2}\left(1 + \sqrt{1+8g^2}\right) \approx1+4g^2  $; so the energy degeneracy is completely lifted. 
However, the eigenvalues of $\hat A$ are -1 for $|-\rangle$; and +1 for both $|+\rangle$ and $|3\rangle$, and hence $|\tilde{+}\rangle$ and $|\tilde{3}\rangle$. So this operator cannot serve to specify the mixing of $|+\rangle$ with $|3\rangle$ effected by the perturbation.  
When it comes to "good" zeroth-order states in the degenerate subspace, of course, $\hat A$ will "prefer" $|+\rangle, |-\rangle$ to $|1\rangle,|2\rangle$, since you see that $\hat A$ characterizes the space in which its degenerate eigenstates mix, so it has made a choice for $|+\rangle$ already--it has lifted the degeneracy, aware of the perturbation. (See Griffiths QM Theorem p. 229, Ch6.)  
For a shortcut, you might look at the linked question. In your example, it is evident $|-\rangle$ does not couple to the perturbation, so it remains unmodified and decoupled; effectively, it drops off the problem.
