Let $F\equiv\{F^a\}_a$ be a POVM in some finite-dimensional Hilbert space $\mathcal X$. It is well-known that one can always understand $F$ as a projective measurement (PVM) in an isometrically enlarged space. More precisely, defining $V_F:\mathcal X\to\mathcal Y\otimes \mathcal X$ as $$V_F u \equiv \sum_a |a\rangle\otimes (\sqrt{F^a}\,u), \qquad u \in\mathcal X$$ one verifies that $V_F$ is an isometry and $F^a=V_F^\dagger (\mathbb{P}_a\otimes I)V_F$ with $\mathbb{P}_a\equiv|a\rangle\!\langle a|$.
Let me now consider a slightly different scenario: let $W:\mathcal X\to\mathcal Y\otimes\mathcal X$ be some isometry (not necessarily tied to any specific POVM). We can now build a corresponding POVM describing evolution through this isometry followed by a PVM on the ancillary space. This reads $$(F_W)^a \equiv W^\dagger (\mathbb{P}_a\otimes I)W.$$ Any isometry can be decomposed wrt a basis for $\mathcal Y$ as $$W u= \sum_a |a\rangle\otimes (W_a u), \qquad u\in\mathcal X,$$ but in general, the operators $W_a\equiv (\langle a|\otimes I)W:\mathcal X\to\mathcal X$ are not positive semidefinite operators. The only thing we can say about them is that, due to the isometric constraint $W^\dagger W=I$, they must satisfy $\sum_a W_a^\dagger W_a=I$ (i.e. they can always be understood as the Kraus operators for some channel).
But then again, given $F_W$, I could follow the construction in Naimark's theorem to obtain another isometry, call it $V_{F_W}$, whish has the form $$V_{F_W} = \sum_a |a\rangle\otimes \sqrt{(F_W)^a}.$$
I'm not entirely sure what to make of this: $F_W$ is clearly a POVM for any isometry $W$, but then it doesn't seem like $V_{F_W}$ equals $W$ in general (as $W_a$ won't necessarily be positive semidefinite, while the corresponding blocks of $V_{F_W}$ are). But still, both isometries feel like they are performing a similar kind of operation. Is there a way to see whether they are indeed "equivalent", at least as far as the measurement at hand is concerned?
