Bounded-index nilpotent subgroups of rational linear groups with finitely many automorphism orbits

Let \\(G\\leq\\operatorname{GL}_n(\\mathbb Q)\\) have finitely many orbits under its full abstract automorphism group, and put \\(d=\\dim_{\\mathbb Q}\\operatorname{span}_{\\mathbb Q}G\\). We prove that \\(G\\) has a torsion-free normal subgroup \\(U\\) satisfying \\[ \\gamma_n(U)=1,\\qquad [G:U]\\leq(2n+1)^d\\leq(2n+1)^{n^2}. \\] Thus \\(G\\) is virtually nilpotent, answering Kourovka Problem 21.40 affirmatively. More generally, if the trace set \\(T=\\{\\operatorname{tr}(g):g\\in G\\}\\) is finite, the same subgroup satisfies \\([G:U]\\leq |T|^d\\). The proof combines bounded-degree roots of algebraic numbers with the trace radical of the rational matrix algebra generated by \\(G\\). Restriction of scalars gives corresponding bounds over number fields.