Given a sequence of real symmetric $(2^N\times 2^N)$- matrices $(H_N)_{N\in\mathbb{N}}$ on the $N$-fold tensor product of $\mathbb{C}^2$ with itself, such that 
\begin{align}
\lim_{N\to\infty}||[H_N,S_N]||_N=0,
\end{align}
where $||\cdot||_N$ denotes the operator norm on $B(\bigotimes_{n=1}^N\mathbb{C}^2)$ (the space of bounded liner operators on $\bigotimes_{n=1}^N\mathbb{C}^2$), $[\cdot,\cdot]$ the usual commutator, and the matrix $S_N$ denotes the symmetrization (projection) operator given by linear extension of the following map on elementary tensors:
\begin{align}S_N (v_1 \otimes \cdots \otimes v_N) = \frac{1}{N!} \sum_{\sigma \in {\cal P}(N)} v_{\sigma(1)} \otimes \cdots \otimes v_{\sigma(N)},
\end{align}
where ${\cal P}(N)$ denotes the symmetric group.
It is known that for all $N\in\mathbb{N}$ the ground state eigenvector $\psi_N$ of $H_N$ (i.e. the eigenvector corresponding to two lowest eigenvalue of $H_N$) is unique and has strictly positive components. My question is if we can conclude that
\begin{align}
\lim_{N\to\infty}||S_N\psi_N-\psi_N||_N=0,
\end{align}
where the latter norm is considered on the space $\bigotimes_{n=1}^N\mathbb{C}^2$.