 Solution

Split the sequence as vector spaces at each vertex. In these coordinates, the middle arrow maps have block formFor $t\in k^{\times }$, change basis by $\mathrm{diag}(t{I}_{X_i},I_{Z_i})$ at every vertex. It changes the off-diagonal block to $t{\xi }_{\rho }$. These arrow matrices depend polynomially on $t$ and at $t=0$ give $X\oplus Z$. Thus the split extension as a degeneration showsWe must check that the two orbits are different. If $Y\cong X\oplus Z$, apply ${\mathrm{Hom}}_Q(Z,-)$ to the original short exact sequence. Exactness makes the kernel of $\mathrm{Hom}(Z,Y)\to \mathrm{End}(Z)$ equal to $\mathrm{Hom}(Z,X)$. The assumed isomorphism of the middle representation gives $\dim \mathrm{Hom}(Z,Y)=\dim \mathrm{Hom}(Z,X)+\dim \mathrm{End}(Z)$, so this map is surjective. In particular the identity of $Z$ lifts to $Z\to Y$, splitting the sequence, a contradiction. Hence the orbits are indeed distinct.

The group $\mathrm{GL}(\mathbf{n})$ is irreducible, so $\overline{{\mathcal{O}}_Y}$ is irreducible. An algebraic-group orbit is locally closed, hence open in its closure; its boundary is a proper closed subset of smaller dimension. The distinct orbit ${\mathcal{O}}_{X\oplus Z}$ lies in that boundary. Therefore