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In the paper Double helix in large large cardinals and iteration of elementary embeddings there are three things mentioned as unknown which I can answer:

[S]everal results known for ordinary supercompactness do not seem to carry over to $n$-fold versions in the same way, e.g., the statement that if $\lambda$ is supercompact, then $\kappa \lt \lambda$ is supercompact iff $V_\lambda \vDash \text{“$\kappa$ is supercompact”}$.

Are “there is an $n$-fold extendible cardinal” and “there is an $n$-fold supercompact cardinal” equiconsistent, for $n\ge2$?

In one direction, theorem 8.3/corollary 8.4 of Sato's paper shows that $n+1$-fold extendible cardinals are $n+1$-fold supercompact (we can alternatively say $n$-fold hyperhuge following Usuba 2017). By this Mathoverflow answer by Gabe Goldberg, theorem 8.3 can be improved to "if $\kappa$ is $n$-fold $\gamma+2$-extendible then it is $n$-fold $\beth_{\kappa+\gamma+1}$-supercompact". Thus the question is about the converse direction.

Are “there is an $n+1$-S-strong cardinal” and “there is an $n$-S-huge cardinal” equiconsistent?

By proposition A.3 of Sato's paper, any $n$-S-strong/$n+1$-fold Shelah cardinal is $n-1$-S-huge (following Perlmutter 2013, $n$-S-huge cardinals can be called $n+1$-fold Shelah for supercompactness). Thus the question is about the converse direction.

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the statement that if $\lambda$ is supercompact, then $\kappa \lt \lambda$ is supercompact iff $V_\lambda \vDash \text{“$\kappa$ is supercompact”}$.

This holds for $n$-fold supercompact cardinals too. If $\lambda$ is 2-fold supercompact (also known as hyperhuge; Usuba 2017) then it is extendible and thus $\Sigma_3$-reflecting and as the $n+1$-fold supercompactness (we can alternatively say $n$-fold hyperhugeness) of $\kappa$ is $\Pi_3$-definable, it is absolute between $V$ and $V_\lambda$.

Are “there is an $n$-fold extendible cardinal” and “there is an $n$-fold supercompact cardinal” equiconsistent, for $n\ge2$?

If $\kappa$ is $n$-fold $\beth_{\kappa+\gamma}$-huperhuge then it is $n+1$-fold $\gamma$-extendible. Thus $\kappa$ is $n$-fold huperhuge iff it is $n+1$-fold extendible. By the proof of theorem 8.5 of Sato's paper, if $\kappa$ is $n$-fold $\beth_{\kappa+\gamma}$-huperhuge, witnessed by an elementary embedding $j : V \to M$, then $V_{j^{n+2}(\kappa)}^{M^{(n+2)}} \vDash \text{"$\kappa$ is $n+1$-fold $\gamma$-extendible"}$. By elementarity of the $n+1$-th iterate of $j(j)$, this reflects to $V_{j(\kappa)}$. The $n+1$-fold $\gamma$-extendibility of $\kappa$ is upward absolute from $V_{j(\kappa)}$ to $V$.

Are “there is an $n+1$-S-strong cardinal” and “there is an $n$-S-huge cardinal” equiconsistent?

A cardinal is a $n+1$-S-strong/$n+2$-fold Shelah cardinal iff it is $n$-S-huge/$n+1$-fold Shelah for supercompactness.

Lemma: $n+2$-fold Shelah cardinals can also be characterized as $n+1$-fold Shelah for extendibility. This follows from the fact that any elementary embedding $j : V_{\beta+1} \to V_{\gamma+1}$ can be extended to an embedding $j : V \to M$ such that $V_\gamma \subset M$, as noted by this Mathoverflow answer by Joel David Hamkins.

Suppose $\kappa$ is $n+1$-fold Shelah for supercompactness. Then it is $n+2$-fold Woodin/$n+1$-fold Vopenka so for any function $f : \kappa \to \kappa$, $n+1$-fold $f$-extendible cardinals $\beta$ (meaning that there is an extendibility embedding $i : V_{f(j^{n}(\beta))} \to V_\gamma$ with critical point $\beta$ such that $\beta$ is closed under $f$ and $i (f\upharpoonright f(j^{n}(\beta))) = f\upharpoonright \gamma$). For a given $f$ define $g$ such that $g(\alpha)$ is the least $\gamma$ such that there is an $f$-extendibility embedding $i : V_{f(j^{n}(\beta))} \to V_\gamma$ with critical point $\beta$, where $\beta$ is the $\alpha$-th $n$-fold $f$-extendible cardinal. Since $\kappa$ is $n+1$-fold Shelah for supercompactness, there is an elementary embedding $j : V \to M$ such that $M^{\beth_{j(g)(\kappa)}} \subset M$. Using the argument of theorem 8.5 of Sato's paper, we can see that $j \upharpoonright V_{j^{n}(j(g)(\kappa))}$ witnesses that $M \vDash \text{"$\kappa$ is $n$-fold $j(g)$-extendible}$. Note that $M^{\beth{j(g)(\kappa)}} \subset M$ implies $H_{\beth{j(g)(\kappa)}} \subset M$. By definition of $g$ and elementarity of $j$, there is an $j(f)$-extendibility embedding $k : V_{f(j^{n}(\kappa))} \to V_{j(g)(\kappa)}$ with critical point $\kappa$. Since $f(\alpha) \lt g(\alpha)$ for every $\alpha$, we have $j(f)(\kappa) \lt (g)(\kappa)$, so $H_{\beth{j(g)(\kappa)}}$ sees that $k$ witnesses that $\kappa$ is $n+1$-fold Shelah for extendibility with respect to $f$.

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