Retractions for completely positive unital maps, and their effect on spectral diameter 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, 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.)
 A: 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, $\Phi$ 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 $\mathbf v = (I_2 \otimes U)\mathbf w$.
