In a frictionless world (e.g. moon), no. The object that started with an initial velocity will always have that additional energy and therefore a higher velocity.

There is of course a height at which the second object could be dropped such that the additional gravitational potential energy matched the gun energy, and that would equalise the velocities.

In a frictional world, I would expect both to come to the terminal velocity asymtotically. I think a bullet is small enough that its TV might be quite low - maybe around 80 mph. So I would expect the dropped bullet and the fired bullet to become very close to the same velocity after falling a kilometre, for instance. A rifle bullet is considered 'spent' at 1000 yards.

I might try measuring the TV of a bullet tomorrow. Here is how. Hang a piece of lead on a thread from my car radio aerial. Drive at an increasing speed until the thread makes an angle of 45 deg with the vertical. At that point, the vectors of gravity and drag on the lead must be equal, which is the same condition as for terminal velocity. So that's how fast I am driving (careful about turbulence and aerodynamics on the vehicle, though, and you don't need the wind to be strong either).

As a thought experiment, I visualise the lead being at 45 deg at quite a slow road speed, so I can still turn the headlights on safely.

I might also try with a dead mouse, with a view to finding out how fast it falls down a well. I will leave the horse for another day.

Been keeping an eye on the questions today? Why does it rain while the sky is blue?