don't confuse youself, condensation takes place on cooling and may occur after compression...

If the two gas samples are truly identical, then it shouldn't make any difference what temperature they start cooling from - all that happens is that one of them is superheated compared to the other.

But if natural gas is dropping condensate or water out at 5C then it's not been very well conditioned...

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Pressure in the hotter pipe will drop to 0.9Po, while that in the colder pipe will drop down to 0.96Po. At this composition and P/T, if we assume retrograde condensation is not likely, there should be slightly more condensate dropout in the 16degC pipe. A process simulation run should tell you what would actually happen; model this as a batch cooling operation.

I think you would need to investigate the thermodynamic path (P, T diagram) of the gas at constant enthalpy from hot to cold condition, given a final pressure after cool down.
How the paths cross the gas two-phase envelope may induce some differences in each case but it should be very minor.

"Main point am trying to ask is does the higher temperature mean there will be more liquid dropout as it has to cool more to reach the ambient temp?"

Since originally, both cases should normally start from outside the two-phase envelope (if conditioning is proper), they would roughly cross same distance through the bell curve. But then again check with my first point for possible differences.

One remark... if the given gas is hotter than the other and pressure is identical in both cases, how could you have the same volume?

At the same pressure but different temperatures the higher temperature gas will be less dense so has less water in it.

That will be a key factor but there won't be a lot in it.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.