Consider a non-singular, completely positive, unital map $\Psi: \mathbf M\_k(\mathbb C) \to \mathbf M\_h(\mathbb C)$. This map will have one or more retractions $\Phi: \mathbf M\_h(\mathbb C) \to \mathbf M\_k(\mathbb C)$.


For such $\Psi$, consider a choice of retraction $\Phi$ and integer $n > 0$. Consider a Hermitian operator $E \in \mathbf M\_h(\mathbb C) \otimes \mathbf M\_n(\mathbb C)$ with unit operator norm, and $\mathbf v, \mathbf w \in \mathbb{C}^{kn}$ be the eigenvectors of $(\Phi \otimes \mathrm{id}\_n)(E)$ with the highest and lowest eigenvalues $\lambda\_{\max}$ and $\lambda\_{\min}$ respectively. Suppose that 
$$ \mathrm{sdiam}\Bigl( \bigl(\Phi \otimes \mathrm{id}\_n\bigr)(E) \Bigr) = \lambda\_{\max} - \lambda\_{\min} > 2 \lVert \Phi \rVert ;$$
does it follow that $\mathbf v = (1 \otimes U)\, \mathbf w$ for some isometry $U \in \mathrm{U}\_n(\mathbb C)$?


For instance: in the case $h = k$ (in which $\Phi = \Psi^{-1}$ would be unique), are there $\Psi$, $n>0$, and $E \in \mathbf M\_h(\mathbb C) \otimes \mathbf M\_n(\mathbb C)$ with unit operator norm, such that 
$$\begin{align}
\lambda\_{\max}\Bigl( \bigl(\Phi \otimes \mathrm{id}\_n\bigr)(E) \Bigr) &> \lVert \Phi \rVert, \tag{1}\\ \quad\text{and}\quad \lambda\_{\min}\Bigl( \bigl(\Phi \otimes \mathrm{id}\_n\bigr)(E) \Bigr) &< - \!\!\;\lVert \Phi \rVert, \tag{2}\end{align} $$
but where the associated eigenvectors $\mathbf v$, $\mathbf w$ of $(\Phi \otimes \mathrm{id}\_n)(E)$ do not merely differ by an isometry on the second tensor factor?


(This question is a follow-up to a [previous question](https://mathoverflow.net/q/285178/3723), in which it was established that there are maps $\Psi$ and operators $E$ for which every retraction $\Phi$ satisfies Eqn. (1) above. Note that I have made a significant revision to this question to exclude a construction which I had intended but failed to exclude, which follows as a corollary to the answer to that question.)



Retractions for completely positive unital maps, and their effect on spectral diameter
Define $\Phi\left(\left[\begin{matrix}a&b\\ c&d \end{matrix}\right]\right) = \left[\begin{matrix}3a - 2d& \frac{5}{2}c\\ \frac{5}{2}b&3d-2a \end{matrix}\right]$. Per [the previous question](https://mathoverflow.net/questions/285178/retractions-for-completely-positive-unital-maps-with-particularly-nice-norms), $\Phi$ is a ucb map that is the retraction of a ucp map. As well, $\|\Phi\| = 5$.


Let $$
E = \left[\begin{matrix} 1\\ &&1 \\ &1 \\ &&&1 \end{matrix}\right] 
\oplus \left[\begin{matrix} -1 \\ &&-1\\ &-1 \\ &&&-\frac{3}{4} \end{matrix}\right] 
\in \mathbf M\_2 \otimes \mathbf M\_4
$$
where this tensor is understood as a 4x4 matrix with 2x2 entries (the direct sum is just for ease of writing). 
Then $E$ is norm 1 and Hermitian. Now
$$
(\Phi\otimes {\rm id}\_4)(E) = \left[\begin{matrix} 3&&&\frac{5}{2}\\ &-2 \\ &&-2 \\ \frac{5}{2}&&&3 \end{matrix}\right] 
\oplus \left[\begin{matrix} -3&&&-\frac{5}{2} \\ &2\\ &&\frac{3}{2} \\ -\frac{5}{2}&&&-\frac{9}{4} \end{matrix}\right]
$$
which gives that
$$
\lambda\_\max((\Phi\otimes {\rm id}\_4)(E)) = \frac{11}{2} > \|\Phi\|
$$
with corresponding eigenvector $\mathbf v = (1,0,0,1,0,0,0,0)^T$ and (if I've done the calculations correctly)
$$
\lambda\_\min((\Phi\otimes {\rm id}\_4)(E)) = -\frac{21 + \sqrt{409}}{8} < -\|\Phi\|
$$
with corresponding eigenvector $\mathbf w = (0,0,0,0, w\_1,0,0,w\_2)^T$. It should be clear that $\left(\begin{matrix} 1\\1\end{matrix}\right)$ and $\left(\begin{matrix} w\_1\\ w\_2\end{matrix}\right)$ are not scalar multiples of each other. Therefore, there is no unitary $U\in \mathbf M\_4$ such that $\mathbf v = (I\_2 \otimes U)\mathbf w$.


