I think you can fix it by taking the derived basechange instead. This corresponds to asking that you're flat on $\mathbb{P}^1_{\mathbb{F}_2}$ and that $2$ acts injectively on the original module. (There's probably a more algebro-geometric way of saying this.) That latter part fails in your case.

Your criterion for checking flatness at the special fiber seems not to work as stated. For example, wouldn't it also show that $\mathcal{O}/2$ is flat on $\mathbb{P}^1_{\mathbb{Z}_{(2)}}$? (or, even easier: wouldn't it apply to $\operatorname{Spec}(\mathbb{Z}_{(2)})$ and show that $\mathbb{Z}/2$ is a flat $\mathbb{Z}_{(2)}$-module?)

Doesn't your example work as a counterexample when you take $k=\mathbb{Z}$? Or does that violate some condition I'm missing?

Property (1) even tells you that the action factors through the abelianisation of $G$ (you can show $(gh)\cdot (\sigma\circ id) = (hg) \cdot (\sigma\circ id) $), so it seems it's closer to the monoidal structure referred to by Zhen Lin.

It seems to me that the compatibility with composition asked for here is NOT the one relating to the standard monoidal structure David explains. Otherwise we would have $g\cdot(\sigma_1\circ \sigma_2) = (g\cdot \sigma_1)\circ (g\cdot \sigma_2)$

I think this "barcode" description relies on field coefficients. A sequence of abelian groups like $0\to \mathbb{Z} \to \mathbb{Z}/2\to \ldots$ certainly doesn't decompose into "bars" that just look like a single generator appearing and disappearing again. You can of course still do homological algebra of $[0,1]$-indexed sequences of abelian groups with finitely many critical values, and homotopy groups of a $[0,1]$-filtered space give an example, but it's unclear what exactly you're asking.

Ah. Then you can still argue that the cohomology groups are finitely generated as $\mathbb{Z}_p$ modules, and since they're also torsion, they are finite abelian p-groups.

If I understand correctly, and lattice means $\mathbb{Z}^n$, then this follows from the fact that cohomology of finite groups with coefficients in finitely generated abelian groups is always finite (since it is torsion and finitely generated)

In your concrete situation, the long exact sequence, together with the fact that cohomology of finite groups with rational coefficients vanishes, should give you what you want.

Small observation: in general, even the action of a monoid on itself isn't necessarily free in this sense. (Only if the monoid is cancellative.) So maybe a good first question is to ask when $\operatorname{End}(1)$ is cancellative?

Just a comment because I don't know what to say about the general case: For the opposite of Fin, we would get a "quotient coclassifier" in Fin, i.e. an epimorphism such that every other epimorphism is uniquely a pushout along it. But that's absurd, for example pushouts along a fixed epimorphism of finite sets can't reduce the cardinality by more than a fixed amount.

You write "whose characteristic does NOT divide the order", is that a typo? I'm asking since you also added the "modular representation theory" tag. Typically modular representation theory refers to the case where the characteristic divides the group order. In characteristic not dividing the group order, the representation theory behaves as in characteristic zero, in particular your representation is irreducible.

$HC$ might be dangerous notation, as it collides with cyclic homology.

You don't even need the precise analysis of the fiber: Any fibration homotopic to the constant map will have fiber $S^3 \times \Omega S^2$, not a manifold. This directly shows that the constant map can't be homotopic to a submersion. (Riemannian is completely irrelevant)