If an ideal current source was connected to the drain of a common source amplifier and nothing else was, i.e. there was no output load then the input voltage could not modify the drain current.
However this scenario quickly breaks down in both ways highlighted above.
The current source is never ideal : however lets assume it does have infinite output impedance. It can still not supply current beyond its own supply rails : Vdd and Gnd.
So what happens when we change the gate voltage? For a gate voltage below threshold, the FET is off (refusing to conduct), Vout = Vdd minus a small drop in the current source, which is saturated, or trying as hard as it can to supply current.
Here an ideal current source would simply raise the voltage until the FET breaks down, in order to pass its current ... actually not a good thing.
As we increase the gate voltage above Vth, the FET starts conducting, until eventually it conducts the rated current : now Vout falls below Vdd. Increase gate voltage a little more and the FET conducts more : Vout quickly falls close to 0 and can fall no more.further. So a tiny increase in Vg produces a very large change in Vout : a current source as a load is a trick to give very high voltage gain (ideally, infinite, like an op-amp. Which is why you'll see this trick used in FET opamps).
In a normal amplifier, there will be negative feedback from output to input, keeping the circuit balanced on that knife edge point between fully on or fully off...
The other way this picture breaks down is in the presence of a load resistance. This acts in parallel with the current source, connected from "OUT" to either Vdd or Gnd. (If it's connected to Gnd, its resistance R should be higher than Vdd/I(currentsource), or it will consume all the current when the FET is off, and limit Vout to I * R).
Now, as the FET turns on, it draws current through the resistor. changing Vout as the FET current changes. Now, given the transconductance and load resistance, you can calculate the finite gain of this circuit.