Given two states $\rho_A, \sigma_A$, Uhlmann's theorem states that the fidelity between them is achieved in the following way
$$F(\rho_A, \sigma_A) = \max_{U_{R'}}F(\rho_{AR'}, (I\otimes U_{R'})\sigma_{AR'})$$
where $U_{R'}$ are unitary operators and $\rho_{AR'}, \sigma_{AR'}$ are arbitrary purifications.
Instead of purifications, let us consider extensions of the state. That is, let $\rho_{AR}$ (possibly a mixed state) satisfy $\text{Tr}_R(\rho_{AR}) = \rho_A$ and $\sigma_{AR}$ (possibly a mixed state) satisfy $\text{Tr}_R(\sigma_{AR}) = \sigma_A$. I am aware of the following fact (5.33 of these notes): For any fixed extension $\rho_{AR}$ we have $$F(\rho_A, \sigma_A) = \max_{\sigma_{AR}}\{F(\rho_{AR}, \sigma_{AR}) : \text{Tr}_R(\sigma_{AR}) = \sigma_A\} \tag{1}$$ Let $\sigma^*_{AR}$ be the state that achieves the maximum above.
Questions
How is $\sigma^*_{AR}$ related to an arbitrary extension $\sigma_{AR}$? If they were purifications, one had the simple fact that they were related by a unitary on $R$.
If $|R|$ was large enough that there exist pure extension states $\sigma_{AR}$, then is $\sigma^*_{AR}$, the extension that achieves the maximum in $(1)$, also pure?