Since heat transfer needs a finite temperature difference, it seems unlikely. However, the enthalpy of steam is greater than water (due to the latent heat of vaporisation) so is there a possibility that the system is not in equilibrium? (Both the water and steam are at one bar)
If both the water vapour and the water liquid were both at 100$^\circ$C and 1 atmosphere when they were brought together there would be no change as the vapour and the liquid would be in equilibrium.
However suppose that the vapour and the liquid were at 100$^\circ$C but the vapour was at a lower pressure. They were brought together in a rigid, thermally insulated container. The vapour and the liquid would not be in equilibrium and some of the liquid would change into vapour. This would continue until an equilibrium state was reached. There would be no change in the internal energy of the vapour-liquid system but the temperature of the system would be lower than 100$^\circ$C as a change of state from liquid to vapour requires energy.
When the liquid and vapour were brought together the rate at which molecules escaped from the liquid would be greater than the rate at which molecules returned to the liquid. Thus the number of molecules in the vapour phase would increase until the rate of escape was equal to the rate of return. A conversion of liquid to vapour represents an increase in potential energy (molecular bonds are broken) which in turn would mean a decrease in the average kinetic energy of the molecules - the temperature would decrease.