There are cases where the self-energy is zero to all orders in perturbation theory, in any dimension $d$, and the fixed point might be Gaussian or not.
Usually, they correspond to theories where there are causality or particle conservation constraints. I will just give one example.
Take non-relativistic interacting bosons at zero chemical potential and zero temperature. Since the system is empty, the exact 2-point function is that of a free particle
$$G(\omega,p)=(i\omega+\frac{p^2}{2m})^{-1},$$
which means that the self-energy vanishes exactly, even if the Hamiltonian is quartic.
The reason is that all possible self-energy diagrams need to have a closed-loop, that exactly vanishes due to causality (this is related to the fact that the inverse propagator is linear in frequency). It is more obvious physically: the system is empty, and thus a single particle propagates freely.
On the other hand, the 4-point function, corresponding to the scattering of two particles, is renormalized. For the coupling constant $g$, the exact RG equation reads
$$ \partial_s g = \epsilon g + C g^2,$$
with $\epsilon = d-2$ and $C$ some constant. One sees that for $d<2$, there is a non-trivial fixed point, whereas the interaction is irrelevant for $d\geq2$. In $d=2$, the interaction vanishes logarithmically. (Of course, in $d=3$, the dimensionful, renormalized, interaction is finite, and related to the s-wave scattering length. We are here talking about the coupling constant in units of the RG scale, which does vanish.)
One can show that indeed, this physics does correspond to a critical point, with scale invariance, but mean-field critical exponent (even for the non-Gaussian fixed point). Is this a critical point between two phases? Yes! It is a quantum critical point between the vacuum (for negative chemical potential) and a superfluid phase.
All this is discussed in S. Sachdev's book "Quantum Phase Transition".