Bounty awarded! Are there any obstructions to trying to get all the listed L-like principles at once?

Does the indestructibility construction give relative consistency of the conjunction of GCH and V=HOD with these large cardinals?

Thank you for the thorough answer! Can you comment on the case of extendible cardinals? I’m always nervous about those due to their high quantifier complexity.

We’ll justify $3 \rightarrow 1$ by the contrapositive. Fix $s \in \prod A_i.$ Identify $\bigsqcup A_i \setminus \{s_j: j<\omega\}$ with $\{t \in \prod A_i: \exists ! j (s_j \neq t_j)\}.$ The latter injects into $\mathbb{R}$ and is thus orderable.

Nitpick: the phrase “Scott cardinals” makes your questions unnecessarily reliant on Foundation. It’s fine here to just have $\kappa, \lambda$ be arbitrary sets.

$\mathcal{P}(X) \cong \mathcal{P}(X^2) \supseteq \{\text{wellorderings of } X \}\ge^* \kappa \Rightarrow \mathcal{P}^2(X) \ge^* \mathcal{P}(\kappa) \ge^* \kappa^+ \Rightarrow \kappa^+ \le \mathcal{P}^3(X).$

@AsafKaragila The condition $X=X^2$ is enough to get $\aleph(X) \le \mathcal{P}^2(X),$ though I did need the third power set for the successor cardinal. I still suspect the claim holds for just a single power set, perhaps by listing out as many comparisons as possible among the cardinalities between $\mathcal{P}^k(X)$ and $\mathcal{P}^{k+1}(X)$ for $k$ up to say 10, and eventually finding a contradiction. I don’t think it would be a particularly enlightening endeavor though.

See mathoverflow.net/a/471784/109573

It also goes through in Z (without foundation).