Here's a diagram of the gauge wiring you mention.

Lowering the voltage brings in the issue of Ohms law which shows the relationship between voltage resistance and current. It turns out that resistance and voltage are directly proportional to current.

I did some calculations and they show this.

with the resistance constant and the voltage drop we get.

V=12 R=45 C=.2666 where V is voltage, R is resistance, and C is current.

We drop the voltage by 1/4th to 9 and we get,

V=9 R=9 and C becomes .2

Dropping to 6 volts gives,

V=6 R=45 and C becomes .1333

So we go to resistance and we see that if we leave the voltage constant and increase the resistance by 1/4 we get,

V=12 R=60 and C becomes .2

If we increase by 1/2 we get,

V=12 R=90 and current becomes .1333.

If we drop the resistance we get a larger increase in current.

A 1/3 drop in resistance gives,

V=12 R=30 current doubles to .4

Another drop in resistance of 1/3 gives

V=12 R= 15 and C becomes .8.

Remember, when we ground the sending unit side of the gauge we get an empty reading on it, so this would seem to bear out the idea that lower resistance and more current will give a lower reading on the gauge.

The temperature gauge is the opposite, when we ground the sender we get a higher reading on the gauge.

My conclusion: Lowering the voltage will not have nearly the effect as lowering the resistance.