Something along the lines of what you want is explained in these lecture notes by Argyres.
As I understand it, the essence of the idea is that it explores the consequence of considering interacting theories in which you have Lorentz symmetries plus "internal" symmetries which are not separated in the usual sense, i.e. the internal symmetry generators don't commute with the Lorentz generators as they do in conventional theories (for example the generator of U(1) gauge transformations conventionally commutes with the generators of Lorentz transformations). The consequence is that the interactions become trivial in the sense that the allowed scattering angles become discrete (and hence vanishing by analyticity of the S matrix).
In a scattering process, the conservation laws resulting from translation invariance and Lorentz invariance leave only the angle of the scattering undetermined. If now, the theory is invariant under a symmetry group that mixes up tranlations, Lorentz transformations and some internal symmetries, then the extra conservation laws place restrictions on the scattering angle, leaving only a discrete allowed set.
Argyres gives the following example:
Take a theory with one of these "mixed up" symmetries - we have a conserved charge $Q_{\mu\nu}$. This is an internal symmetry generator which also has Lorentz indices. It's assumed to be symmetric and traceless (presumably for irreducibility). Its matrix element in a one-particle momentum eigenstate is of the form $$\langle p|Q_{\mu\nu}|p\rangle \propto p_{\mu}p_{\nu}-\frac{1}{4}\eta_{\mu\nu}p^2$$
It is reasonable to assume that in a two particle state, its matrix element is of the form $$\langle p^1p^2|Q_{\mu\nu}|p^1p^2\rangle = \langle p^1|Q_{\mu\nu}|p^1\rangle+\langle p^2|Q_{\mu\nu}|p^2\rangle $$ Considering now a process where a pair of momenta $p^1, p^2$ elastically scatter into $q^1, q^2$ then if $Q_{\mu\nu}$ is conserved $$\langle p^1p^2|Q_{\mu\nu}|p^1p^2\rangle=\langle q^1q^2|Q_{\mu\nu}|q^1q^2\rangle $$ $$\langle p^1|Q_{\mu\nu}|p^1\rangle+\langle p^2|Q_{\mu\nu}|p^2\rangle = \langle q^1|Q_{\mu\nu}|q^1\rangle+\langle q^2|Q_{\mu\nu}|q^2\rangle$$ $$p^1_{\mu}p^1_{\nu}-\frac{1}{4}\eta_{\mu\nu}(p^1)^2+p^2_{\mu}p^2_{\nu}-\frac{1}{4}\eta_{\mu\nu}(p^2)^2 = q^1_{\mu}q^1_{\nu}-\frac{1}{4}\eta_{\mu\nu}(q^1)^2+q^2_{\mu}q^2_{\nu}-\frac{1}{4}\eta_{\mu\nu}(q^2)^2 $$ Hence $$ p^1_{\mu}p^1_{\nu}+p^2_{\mu}p^2_{\nu} = q^1_{\mu}q^1_{\nu}+q^2_{\mu}q^2_{\nu} \ \ (1)$$ Four momentum is also conserved $$p^1_{\mu}+p^2_{\mu} = q^1_{\mu}+q^2_{\mu} $$ Now suppose scattering changes the momenta by $$q^1_{\mu} = p^1_{\mu}+a_{\mu}; \ \ \ q^2_{\mu} = p^2_{\mu}+b_{\mu}$$ Four momentum conservation tells us that $b_{\mu} = -a_{\mu}$. Substituting for the $q$'s in (1) we eventually get $$a_{\mu}(p^1_{\nu}-p^2_{\nu})+ a_{\nu}(p^1_{\mu}-p^2_{\mu})+2a_{\mu}a_{\nu} = 0$$ Since this holds for all values of $p^1_{\mu}, p^2_{\mu}$ we deduce that $a_{\mu}$ must vanish, i.e the interaction is trivial in the sense that $q^1_{\mu} = p^1_{\mu}$ and $q^2_{\mu} = p^2_{\mu}$ i.e. the scattering angle is zero.
