That's not really possible. To see why simply express the Gibbs' free energy for each of the phases of the homogeneous substance in differential form:
$$dG = Vdp - SdT$$
We assume constant composition.
We can identify the volume $V$ as the partial derivative of the free energy with respect to pressure at constant temperature:
$$V = \left( \frac {\partial G}{\partial p}\right)_T$$
Since volumes are always positive quantities, the general response of a pure substance at constant temperature to a decrease in pressure is a decrease in free energy ($(dG)_T = Vdp$). The only way we can have a solid with a lower free energy than the corresponding gas as we lower pressure is therefore if there is a discontinuity (break from monotonous behavior) marked by a sudden increase in the free energy at the transition point from solid to gas. But since $V$ is, by definition, greater in a gas than in a condensed phase such as a solid, the free energy change at the transition point from gas to solid is $(d \Delta G)_T = \Delta Vdp >0$ if we decrease the pressure ($dp<0$), so that the solid is identified as the less stable phase.