My question is about an isomorphism between two varieties of the form discussed in this thread; it is also related to an earlier question of mine.

Let $z_n$ be the standard nilpotent of type $(n,n)$. For $1 \leq k \leq 2n-1$, let $P_k$ be the parabolic subgroup of $GL(2n)$ stabilizing a flag of the form $(0 \subset V_1 \subset \cdots \subset \widehat{V_k} \subset \cdots \subset V_{2n}=V)$. Let $S_n$ denote the Slodowy slice to $z_n$, and let $\pi_k: T^* G/P_k \rightarrow \mathfrak{gl}_{2n}$ be the natural projection. Consider also the Borel $B \subset GL(2n)$, and the Springer map $\pi: T^*G/B \rightarrow \mathfrak{gl}_{2n}$. Then my question is: Why is $\pi_k^{-1}(S_{n}) \simeq \pi^{-1}(S_{n-1})?$

This claim is made on page 10 (immediately after the proof of Lemma $5$) in this paper. I understand why $\pi_k^{-1}(z_n) \simeq \pi^{-1}(z_{n-1})$ (this is the Lemma $5$ I just mentioned). This follows, since if $(0 \subset V_1 \subset \cdots \subset \widehat{V_k} \subset \cdots \subset V_{2n}=V) \in \pi_k^{-1}(z_n)$, it is stabilized by $z_n$, then $z_n V_{k+1}=V_{k-1}$ since $\text{ker}(z_n)$ is $2$-dimensional; and we can map this flag to $(0 \subset V_1 \subset \cdots \subset V_{k-1}=z_n V_{k+1} \subset \cdots \subset z_n V_{2n}) \in \pi^{-1}(z_{n-1})$. A similar argument may be made when we replace $z_n$ by a nilpotent $x \in S_n \cap \text{im}(\pi_k)$, showing that $\pi_k^{-1}(x) \simeq \pi^{-1}(x|_{xV})$ but then the vector space $xV$ is not a fixed vector space, and I am having trouble patching up the isomorphism.

I hope it's ok for me to post this question here (maybe it's better to ask the author of the paper, but I wasn't able to contact the author in this case.)