Let the stabilized channel fidelity between two channels $M_{A\rightarrow B}$ and $N_{A\rightarrow B}$ be defined as
$$F(M,N) = \min\limits_{\vert\psi\rangle_{AR}} F\left((M\otimes I_R)\vert\psi\rangle\langle\psi\vert, (N\otimes I_R)\vert\psi\rangle\langle\psi\vert\right)$$
Let $U_{A\rightarrow BE}$ and $V_{A\rightarrow BE}$ be some Stinespring isometries for $M$ and $N$ respectively. Then we have
$$\begin{align} F(M,N) &= \min\limits_{\vert\psi\rangle_{AR}} F\left((M\otimes I_R)\vert\psi\rangle\langle\psi\vert, (N\otimes I_R)\vert\psi\rangle\langle\psi\vert\right) \\ &=\min\limits_{\vert\psi\rangle_{AR}} \max_{V} F\left((U\otimes I_R)\vert\psi\rangle\langle\psi\vert(U^\dagger\otimes I_R), (V\otimes I_R)\vert\psi\rangle\langle\psi\vert(V^\dagger\otimes I_R)\right)\\ &\geq \max_{V} \min\limits_{\vert\psi\rangle_{AR}} F\left((U\otimes I_R) \vert\psi\rangle\langle\psi\vert(U^\dagger\otimes I_R), (V\otimes I_R)\vert\psi\rangle\langle\psi\vert(V^\dagger\otimes I_R)\right) \\ &= \max\limits_{V} F(U(\cdot)U^\dagger, V(\cdot)V^\dagger), \end{align}$$
where the second line uses Uhlmann's theorem to pick a specific $U$ and we maximize over all $V$, the third line is the maxmin inequality and the last line uses the definition of channel fidelity.
I would like the reverse bound as well. That is, can we show that the channel fidelity $F(M,N)$ is upper bounded by the channel fidelity of the channel corresponding to some Stinespring isometries of $M$ and $N$? Equivalently, can we swap the min and the max with equality somehow in the above argument?
The analogous result for states is $F(\rho, \sigma) = \max_{\phi} F(\psi,\phi)$, where $\psi$ is a specific purification of $\rho$ and we maximize over all $\phi$ which are purifications of $\sigma$. That is, of course, Uhlmann's theorem.
