Welcome to Mathoverflow! Are you asking about isomorphism as abelian groups or as rings? I think these groups will be free of the same rank (equal to the number of maximal cones in the fan) but not in general isomorphic as rings.

Are you sure about the ag.algebraic-geometry tag? Seems a bit far away!

Note that if you replace $i$ by more general proper morphism, e.g. resolution $\pi: Y \to X$ of rational singularities this is presumably unknown: according to Kawamata Lemma 7.4 in arxiv.org/pdf/1903.00801.pdf, the image of $\pi_*$ generates the $D^b(X)$ up to direct summands, but without direct summands this is not known (at least not in the literature and I don't know how to prove it).

@naf: thanks, I will try to work out the details of the argument. I've tracked the historical development of this subject: Shafarevich 1962 -> Parshin 1968 -> Arakelov 1971 -> Faltings 1983 -> Deligne 1987. The original paper by Shafarevich is his ICM 1962 talk in Russian mathunion.org/fileadmin/ICM/Proceedings/ICM1962.1/… (page 174), where he says the result is true for hyperelliptic curves of g > 1.

@naf: thanks, if you'd like to, please feel free to put this as an answer.

@JasonStarr: I clarified the question, hopefully it's clearer now. I thought about your suggestion of using Vakil's work on rational ellitpic surfaces but I don't see the complete argument: so rational elliptic surfaces are determined by their branch points, and hence possibly elliptic K3s which are double covers of those too, but what if you have more K3s with the same branch points which are not double covers??

Thanks Jason! Discriminant is arbitrary, I consider rank two elliptic K3s as e.g. in Section 3.2 of arxiv.org/pdf/1907.01335.pdf.

Any smooth projective toric variety is a compactification of $\mathbf{C}^n$, because its toric open chart corresponding to a maximal cone will be affine smooth toric variety without torus factors, hence isomorphic $\mathbf{C}^n$.

Derived categories of coherent sheaves vs derived category of quiver representations makes the analogy more precise. "Genus of a quiver" should probably be thought as zero as the Grothendieck group of $\mathrm{Rep}(Q)$ has no continuous part so "noncommutative Jacobian" is trivial. The other questions do not seem to allow for immediate categorical interpretation.