It will be obvious on consideration that
- "An LED drawing 20 mA when you applied 2.2V would draw 20 mA when you applied 2.2 Volts" :-)
ie in you ideal situation you have stability. But your question shows that you are well aware that in the real world this ideal situation would not apply. Realising this is a good start.
As Starblue said, looking at the datasheet is a good idea.
Below is the voltage versus current curve for a reasonably typical LED. The makers say that it nominally is rated at 100 mA at 3.2V but a look at the curve show that as drawn it drops 3.3V at 100 mA. The table and the curve are both meant to be "typical" values - it's a bad start when a datasheet disagrees with itself even slightly, but it's unimportant here an demonstrates the inexact relationship between Vf and If in practice for a randomly chosen LED. The data sheet says Vf could be as low as 2.9V and as high as 3.5V at 100 mA.
Look at the curve and note what happens for a typical LED if Vf is changed from 3.3V (=100 mA) to 3.4V. At 3.7V it draws 200 mA, at about 4.05V = 300 mA and at 4.4V it draws 400 mA.
That is, for a change of Vf from 3.3V to 4.4V =~ a 33% increase in Voltage, current goes from 100 mA to 400 mA.
A few simple exercises, which will greatly help your understanding if you do them yourself, will give you a much better feel for what happens in real life. Try working out the stable state on this curve for a fixed power supply voltage and a series of resistor values. Then try a fixed resistor value and a series of power supply voltages.
Tell us if this helps your understanding.