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This shows it depends on p, but at least for the features you mention in your answer, it seems to depend on p in a very uniform way. (Analogous, in my mind, to the classification of finite groups of order $p^7$; obv. the groups depend on $p$, but the classification is "independent" of $p$ for $p \geq 7$.) Of course, Cam's answer and the Boston-Ellenberg paper from the comments undermine even this kind of uniformity.
You don't need a Euclidean ring (nor Noetherian!) to get Hermite normal form, just a Hermite ring (Kaplansky, "Elementary Divisors and Modules", Trans. AMS, 1949). I believe it is almost trivial to show that your ring $B$ is Hermite: you just need that for any $(a,b)$, there is an invertible 2x2 matrix $Q$ over $B$ such that $(a,b)Q = (d,0)$ for some $d$. The key thing to show is that this is true for quasipolynomials; I think the trichotomousness (trichotomosity?) almost follows for free if $a,b$ are.
Connected components is a potentially interesting question, sure. FYI, it need not be the same as # of isolated orbits, e.g. consider the action of the circle by rotations on a cylinder: 1 connected component, no isolated orbits. Given that your equations are polynomials, another standard question to ask is # of irreducible components. You can then ask about which ones intersect, the structure of their intersections, their singular loci, dimensions of all of these objects, etc.
Yes. Your group is connected, and every orbit is the image of the group under a polynomial map (namely, the orbit of a point p is the image of the map $g \mapsto g.p$), so every orbit is connected.
Sorry, I meant the second part of my comment applies: with such a symmetry group, the only possible isolated point is the origin. A more interesting question might thus be to ask about isolated orbits, but I suppose that depends on your motivation.
I don't see why the group action preserves the solution set. E.g. consider the term $X^T Y W_1$ in the first equation... But if you did have such a group action, the only possible isolated point is the origin, since that's the only fixed point of this action.
I don't know whether this specific question was open, but it is certainly in the family of open problems highlighted by Hilbert's 2nd Problem and the Entscheidungsproblem.
There's a central extension $N \hookrightarrow H \twoheadrightarrow G$, where N is the f.g. Abelian group generated by the image of the cocycle, such that $K_\alpha G$ is a quotient of $KH$. However, knowing which simple factors of $KH$ get killed by this quotient seems difficult in general (though perhaps quite feasible in any given example).
@Joel David Hamkins: why not add an answer covering some of those then? Each answer need not be a complete list (the question is even CW). In particular, I'd especially love to hear your take on open questions in Borel equivalence relations, and the modal logic of forcing.
@Benjamin Steinberg: semigroup isomorphism from multiplication tables is equivalent to graph isomorphism under polynomial time reductions. Suggests that it is hard in general, but perhaps good graph iso software (nauty, traces, bliss, conauto) could be of use for particular instances.
