$\DeclareMathOperator\Spf{Spf}\DeclareMathOperator\Pic{Pic}\DeclareMathOperator\Mod{Mod}$One can deduce the invertibility of the Gross-Hopkins dualizing spectrum from purely algebro-geometric considerations. Suppose $G$ is a finite subgroup of $G_n$ (the argument also works for $G_n$ itself); then by Theorem 5.4.4 of Rognes' memoir (see notes/memo0898.pdf), we get that $L_K(n)(\Mod(E^{hG})) = (L_K(n) \Mod(E))^{hG}$. Since $\Spf E$ is self-dual (i.e., the structure sheaf is a dualizing sheaf), it follows that $\Spf E/G$ is self-dual. Because dualizing sheaves are unique up to invertible sheaves, we get that $I_Z E^{hG}$ is in $\Pic(E^{hG})$, so to prove Anderson self-duality, it suffices to show that $\Pic(E^{hG})$ is cyclic. This approach doesn't, of course, tell us anything about what the shift is; but the result's interesting enough without that.
From a similar argument, one can deduce the $K(n)$-local Spanier-Whitehead self-duality of $E^{hG}$ (at $n=p-1$). Since $I_Z = I_Z L_{K(n)} S$ is $K(n)$-locally invertible by Gross-Hopkins duality (see above), we know that $DE^{hG} = I_Z^{-1} \wedge I_Z E^{hG}$. If we know that $I_Z \wedge E^{hG}$ is a shift of $E^{hG}$, we're done. This follows since $I_Z \wedge E^{hG}$ is in $\Pic(E^{hG})$, simply because $(I_Z \wedge E^{hG}) \wedge_{E^{hG}} R = I_Z \wedge R$, and we know that $I_Z$ is $K(n)$-locally invertible.
To actually recover the harder part of Gross-Hopkins duality, we need to identify what $f^! I_Z$ is. Here, I'm writing $f: \Spf E/G_n \to \operatorname{Spec} S$. Recall that $G_n$ is a $p$-adic Lie group of dimension $n^2$ over $Z_p$. Its Lie algebra, $g_n$, is therefore of dimension $n^2$. This implies that $V := pg_n$, the quotient of $g_n$ by the scalars, is of dimension $n^2-1$. One might therefore suspect that $S^V = f^! I_Z$, where $S^V$ is the "one-point compactification" of $V$.
How might one go about proving this, if all this makes sense?
