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@ Will Sawin , I think there is a problem with you argument. Let us say we embedd $f : G \rightarrow GL_n$. Now we know the image of the composite map $f\circ \rho$ lies in $B_u$ (The unipotent radical of a Borel). Set $U$ to be $B_u \cap G$. Now why should $U$ be connected? If it is connected then your argument works. else we will have to work with the connected component of $U$, but then why should it contain $im(\rho)$?
Someone pointed out to me the paper by Borel,Tits in inventiones, where they prove that a subgroup $H \subset G $ consisting of only unipotent elements with $G$ reductive can be realised inside the unipotent radical of a parabolic subgroup $P$ of $G$. I haven't read the paper. Hence I don't really know about the issues involved. But Does assuming the group $H$ being finite make the problem any easier?
