It seems that you are working with the usual (non-derived) dual. In this case, isn't $I_Z^\vee=O_X$, so it's independent of Z?

Well, the Euler characteristic of a slight perturbation can be recovered from the sheaf of nearby cycles of the perturbation (as its Euler characteristic), so the intuition is not really wrong. However, different perturbations might have different nearby cycles sheaves...

P.S. Come to think of it, the morphism need not be injective.

Yes, for any $G$-equivariant sheaf $F$ on $X$, you can consider $F':=(\pi_*(F))^G$ (invariants in the direct image), it is a sheaf on $X//G$. If $F$ descends, then $F'$ is its descent: $F\simeq\pi^*F'$. In general, there is a morphism $\pi^*F'\to F$; it is injective, but not surjective.

End(E)^G makes no sense as a sheaf on X: G-invariance is not defined for local sections when G acts non-trivially on X.

Why do you say that End(E) descends? Are you assuming that the stabilizers of points act trivially (or at least by scalars) on E's fibers?

Let me make sure I understand the question. You basically have two connections on Spec C[t_1,t_1^{-1}], and you are asking when they are isomorphic (or, equivalently, when two matrices over Laurent polynomials are gauge-equivalent), correct? If so, I don't see how to give a meaningful answer: you can use irregular Riemann-Hilbert to describe connections (in terms of monodromy and Stokes data), but this is pretty much impossible to express in terms of $A_1$ and $A_2$.

You probably mean $\{x,y,x+y+z\}$, right?

Thanks for the reference! (Still wondering why the claim was attributed to Deligne...) I will be happy to discuss the statement/proof, but I'd rather do it over email if you don't mind --- it's easier to just send the file.

Thanks, of course I assume that C is essentially small (forgot to put it in the question, now corrected).

Yes, I realize this, and it is also replacing fundamental geometric reason by something more calculation-y, but I thought it might still be worth mentioning as a possible source of intuition.

P.S. Note that if you know the theory of representations of a reductive group, the fact that for any $G$-representation $W$, we have $W^G=W^P$ is almost obvious. Indeed, it is enough to assume that $W$ is irreducible, and then we have to show that $W^P=0$ only if $W$ is the trivial representation. But any vector in $W^P$ is clearly a highest weight vector (because it is invariant under the maximal unipotent), and its weight is zero (because it is invariant under torus), therefore, an irreducible $W$ has $W^P\ne 0$ only if $W$ is the trivial representation.

We do not decompose $V$ into orbits, we just write the space of functions on it as a union of finite-dimensional representations. This is particularly easy if $V$ is a vector space (in your example, $V=\mathfrak{g}$): $\mathbb{C}[V]$ is the union over $d$ of the spaces of polynomials of degree no more than $d$, and each of these spaces is finite-dimensional. (Of course, you can also write it as the direct sum of the spaces of homogeneous polynomials of degree $d$.)