Given the same gain on 2 pentode preamp tubes, would the transconductance ...
It don't work that way.

I haven't experimented with pentodes much, but my sense of it is that if you arrive at the same gain, other issues should be similar. This assumes you don't do something silly, like make a gain stage that has the gain but doesn't have the needed output voltage swing, so the stage is always clipping when you apply an input. But that's a different topic...
The mu of a triode (sometimes called "amplification factor") is the ratio of plate voltage change divided by the grid voltage change that caused it. When you figure the gain of a triode stage, the easiest method is to take the tube's mu and assume that the plate load resistance and internal plate resistance form a voltage divider which reduces the gain possible (mu) to the value actually seen at the plate output. The portion of the output across the plate load is what gets passed along to the next stage.
Mu is also used in triode calculations because of the three basic tube characteristics (mu, transconductance [Gm], internal plate resistance), it is the most-constant parameter. The relationship of the three is such that as plate resistance changes in one direction and Gm changes in the other, they offset and mu changes very little.
But because pentodes have generally horizontal curves for plate current change with increasing plate voltage, it is often assumed their plate resistance is infinite. If you place a load resistor in series with an infinite resistance to form a voltage divider, you get nothing across your load resistor unless it's infinite also. But we get signal out of pentodes, so we need a different model to figure gain.
As an aside, just try to find a pentode data sheet with mu listed. You
might get a rating for mu from g1 to g2, but you won't find one for mu from g1 to plate (which is what triode mu represents). Plate resistance is very high and often difficult to accurately calculate, but is the most constant parameter. This implies mu from g1 to plate will vary with Gm, so the easiest path is to figure Gm from a datasheet at the operating point chosen.
Gm is defined as the change of plate current given a small change of grid voltage. If we assume plate resistance is very high, we
could ignore it, and assume that if we apply a 1v signal at the grid, there is a certain plate current variation that results. So the output will be that plate current change times the plate load resistance, which follows ohm's law (Voltage = Current*Resistance).
End result? If you make Gm bigger, or the plate load resistor bigger, you get a bigger output voltage and therefore more gain.
The problem is if you make the plate load resistor too big (say, approaching infinity), you get very little tube current and Gm drops. So there is a limit to how far you can go by simply making the load resistor bigger before you have to select a different pentode with naturally more Gm. Also, you're probably looking at using a load resistor bigger than you might with a triode, which means the grid reference resistor of the following gain stage has a bigger impact on possibly dropping the gain of the pentode.
Without explaining how the term "micromho" comes about, it is probably helpful to remind you the european method of describing Gm: 1000 micromhos = 1mA/V. That means for a 1v input on the grid, you get 1mA of plate current change. 2,200 micromhos = 2.2mA/V, etc.