we will use a trick. We focus on the $U(N)$ vector model for simplicity, but the argument can be generalized to $O(N)$ as well.

Consider the ${\ensuremath{\mathcal{N}}}=2$ supersymmetric generalization of our theory, which is the ${{\ensuremath{\mathcal{N}}}=2}$ supersymmetric Chern-Simons $U(N)$ gauge theory, coupled to a single matter chiral superfield $\Phi_i$ with components $(\phi_i,\psi_i)$ in the fundamental representation (we use $i,j = 1,\ldots,N$ to label the fundamental representation of $U(N)$).[^4] We will relate $\beta_{\lambda_6}$ in our theory (for infinite $N$) to the beta function of the $(\phi^\dagger\phi)^3$ coupling in the ${\ensuremath{\mathcal{N}}}=2$ theory. The action of the ${\ensuremath{\mathcal{N}}}=2$ theory [@Schwarz:2004yj; @Gaiotto:2007qi], after integrating out all the auxiliary fields, is $$\begin{aligned}
\label{eq:SUSicCS}
S^{{\ensuremath{\mathcal{N}}}=2}_{CS} &= -\frac{ik}{4\pi}\int \operatorname{Tr}\left[A\wedge dA + \frac{2}{3}A^3\right] + \notag\\
&\quad \int \! d^3x \, \left[ |{\ensuremath{\mathcal{D}}}_{\mu}\phi_i|^2 +i\bar{\psi}^i\gamma^{\mu}{\ensuremath{\mathcal{D}}}_{\mu}\psi_i - 2\frac{\lambda}{N}\bar{\phi}^i\phi_i\bar{\psi}^j\psi_j - \frac{\lambda}{N}\bar{\phi}^i\phi_j\bar{\psi}^j\psi_i - \frac{\lambda^2}{N^2}(\bar{\phi}^i\phi_i)^3 \right]
{\,.}\end{aligned}$$ It was shown in [@Gaiotto:2007qi] that this action is exactly conformal quantum mechanically, for all values of $k$ and $N$ (with $\lambda=4\pi N/k$). In particular, this means that the beta function of the $(\phi^\dagger\phi)^3$ coupling in this theory vanishes identically to all orders in $\lambda$ and $1/N$. The theory has a global $U(1)_f$ symmetry acting on the matter superfield as $\Phi \to e^{i\alpha}\Phi$, and in [@Gaiotto:2007qi] it 