I don’t know if this addresses your central question, but the spectral sequences approach to diagram chasing seems relevant here. Though I only know how to do this for diagrams on a square grid, if you have your starting hypothesis, it shows you (all of the?) nontrivial diagram chasing constructed maps as higher differentials. This lets you easily see what exact sequences/isomorphisms come out of different starting exactness hypotheses on your diagram, making it feel more like a mechanical problem.

I think such a base change statement should hold, as open pullback is exact, so preserves images, and one can apply your open pullback to the morphism $j_!$ to $j_*$, and use base change.

The usual identification of symmetric powers of the line with projective space goes via encoding polynomials as their roots, which seems to mess with the point counting equalities you gave. Someone more knowledgeable can hopefully see what’s going on here.

It is! I don’t have a reference off hand, but one can prove it without much trouble using the six functor formalism and that duality respects external products.

The constructible category of sheaves on a singular variety might be an example, there doesn’t seem to be obvious dualising maps for the dualising sheaf.

Over $\mathbb{Z}_p$ I think the question has an affirmative answer, since the category of $\mathbb{Z}_p$ representations of $H_1\cong H_2$ is Krull-Schmidt, so adding copies of the trivial rep doesn’t change isomorphism class. One could maybe do something local/global from this..

Thank you for this example, it was really helpful!

By a generators and relations argument I would think some symbolic argument that holds in any torsion group of exponent coprime to $n$, building $a,b$ from $g,h$ using these assumptions and the assumed $n$th root function. If one could give a group of this type where the result fails, it would rule out such an argument, which would also be very interesting.

The Frobenius group of order $20$ is an example, trivial outer automorphism group, with nonrational characters, take $n=3$.