I maintain that this isn't an absolute fact but is probably true under the outside conditions you are talking about. We should all agree that wind blowing does, in fact, increase the thermal coupling between the thermometer and the air. However, the device has no heat production. If there is no heat production, it doesn't matter how strong or weak the thermal connection between two isolated items is because they will eventually equilibrate to the same temperature.
A major caveat to that is that fact that the device and the air are not in an isolated system. So what else is there? Well, there are heat potential sources/sinks that exist at different temperatures. These include the night sky for instance, the sun, and the ground. There exists radiative heat transfer between all of these and they all constitute a different temperature than the air. To summarize in an equation, there is no heat production, $\dot{Q}$, but there is a heat flow, denoted by $\dot{q}_c(..)$ for conductive and $\dot{q}_r(...)$ for radiative, from everything that can thermally interact.
$$\dot{Q} = 0 = \dot{q}_c(T_\text{thermometer},T_\text{air})+\dot{q}_r(T_\text{thermometer},T_\text{sky})+\dot{q}_r(T_{\text{thermometer}},T_\text{ground})$$
And basically $q_c$ dominates, which could be written with Newton's law of cooling, proportional to $\Delta T=T_\text{thermometer},T_\text{air}$, which means that $T_{\text{thermometer}}=T_\text{air}$ ultimately.