The diamond distance between two channels $\Phi_0$ and $\Phi_1$ is defined in this answer.
$$ \| \Phi_0 - \Phi_1 \|_{\diamond} = \sup_{\rho} \: \| (\Phi_0 \otimes \operatorname{Id}_k)(\rho) - (\Phi_1 \otimes \operatorname{Id}_k)(\rho) \|_1 $$ where $\operatorname{Id}_k$ denotes the identity channel from $M_k(\mathbb{C})$ (the set of $k\times k$ complex matrices) to itself, $\| \cdot \|_1$ denotes the trace norm, and the supremum is taken over all $k \geq 1$ and all density matrices $\rho$ chosen from $M_{nk}(\mathbb{C}) = M_n(\mathbb{C}) \otimes M_{k}(\mathbb{C})$.
Let $N_1$ and $N_2$ be two completely positive trace nonincreasing maps that satisfy
$$\|N_1 - N_2\|_\diamond\leq \varepsilon.$$
For any channel $N_{A\rightarrow B}$, we define its Stinespring dilation to be an isometry $V_{A\rightarrow BE}$ such that $\text{Tr}_E(V\rho V^\dagger) = N(\rho)$.
Can one show that there exist Stinespring dilations $V_1$ and $V_2$ of $N_1$ and $N_2$ respectively such that we also have a bound on
$$\|V_1 - V_2\|_\diamond$$
in terms of $\varepsilon$?