= Solution
The operator $\dot\phi^2\phi$ is not invariant under a <scalar-field shift symmetry> because one field is undifferentiated. An arbitrary large coefficient would therefore radiatively generate equally unsuppressed nonderivative operators, including a mass and a steep <scalar potential>, and would spoil the approximate shift symmetry and flat potential needed for <single-field slow-roll inflation>. In a technically natural slow-roll model its coefficient must consequently be small, so it does not produce parametrically large <primordial non-Gaussianity>. Treated merely as an effective spectator-field interaction, it does generate the tree-level <primordial bispectrum> computed below, whose size is controlled by the dimensionless interaction strength at the Hubble scale; taking that strength large abandons the controlled slow-roll premise.
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