Can any channel be written as $Phi(X)=operatorname{Tr}_{mathcal Z}[U(Xotimes sigma)U^dagger]$ for any state $sigma$?

We know that every CPTP map $Phi:mathcal Xtomathcal Y$ can be represented via an isometry $U:mathcal Xotimesmathcal Ztomathcal Yotimesmathcal Z$, as
$$Phi(X) = operatorname{Tr}_{mathcal Z}[U(Xotimes E_{0,0})U^dagger],quadtext{where}quad E_{a,b}equiv lvert arangle!langle brvert.tag1$$
Showing this is quite easy e.g. from the Kraus representation.
If $A_a:mathcal Xtomathcal Y$ are Kraus operators for $Phi$, then
$$U_{alpha a,i0} equiv langle alpha,arvert Ulvert i,0rangle = langle alpharvert A_alvert i rangle equiv (A_a)_{alpha,0}.tag2$$
We can, of course, replace $E_{0,0}$ with any pure state in (1) without affecting the result.

This shows that, given any channel $Phi$ and any pure state $lvertpsirangleinmathcal Z$, we can represent $Phi$ as in (1) (with $E_{0,0}tolvertpsirangle$).
However, what about the more general case of $E_{0,0}tosigma$ with $sigma$ not pure?

To analyse this case, consider a channel written as
$$Phi(X)=operatorname{Tr}_{mathcal Z}[U(Xotimes sigma)U^dagger]tag3$$
for some state $sigma=sum_k p_k E_{k,k}inmathrm{Lin}(mathcal Z)$ (appropriately choosing the computational basis for $mathcal Z$).
The relation with the Kraus operators reads in this case
$$Phi(X)_{alpha,beta} = sum_{ell k ij} p_ell U_{alpha ell,i k} X_{ij} U^*_{betaell,jk}
= sum_{ell,k} (A_{ell,k}XA_{ell,k}^dagger)_{alphabeta}tag4$$
with
$$(A_{ell,k})_{alpha,i} equiv sqrt{p_ell} U_{alphaell,ik},
qquad A_{ell,k} = sqrt{p_ell} (Iotimes langle ellrvert)U(Iotimes lvert krangle).tag5$$

There is now a difference: the number of Kraus operators must be larger than the rank of $sigma$ (which determines the number elements spanned by the index $k$ in $A_{ell,k}$).
Indeed, a different way to state this same fact is to notice that the $Phi$ in (3) is a convex combination of several channels:
$$Phi(X) = sum_k p_k Phi_k(X), qquad Phi_k(X)equiv operatorname{Tr}_{mathcal Z}[U(Xotimes E_{k,k})U^dagger]. tag 6$$

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This leads me to the question: can any $Phi$ be written as in (3) for any $sigma$? More precisely, given $Phi$ and $sigma$, can I always find an isometry $U$ such that (3) holds?

The question arises from the fact that, because when $sigma$ is not pure (3) leads to $Phi$ be a convex combination of other maps as shown in (6), I would think that there should be maps that are "extremal", in the sense that they cannot be written as convex combinations of other maps, and that such maps shouldn’t be writable as (3) for $sigma$ not pure.