Initial answer
We can answer this extremely interesting (and fundamental for circuit theory) question by conducting thought, real, or simulation experiments with some negative feedback amplifying stage, e.g. an inverting amplifier.
Implementation
To see how the amplifier reaches the equilibrium, we can implement it not by a perfect op-amp with huge open-loop gain but by an amplifier with a relatively small and adjustable gain K (it can be another inverting amplifier with potentiometer but implemented by a perfect op-amp). So this amplifier will be an imperfect inverting amplifier. Instead two discrete resistors R1 and R2, we can use a linear potentiometer to realize the feedback network. Thus we can observe the voltage distribution along its resistive film and visualize it by the help of a voltage diagram. In this geometrical representation, local voltages on the resistive film are represented by vertical bars in red or by their outline (see more about voltage diagram in this Wikibooks story).
Operation
The most important prerequisite for intuitive understanding this phenomenon (called negative feedback) is to think of the amplifier not as of a non-inertial, high speed device (as it is usually presented) but as of some inertial device like an integrator... or even as of some lazy human being who thinks slowly (as me:) This seems strange, but it is extremely important for the intuitive understanding negative feedback circuits.
VIN = 0 V (initial state): Set the potentiometer slider in the middle. Imagine the input voltage VIN is zero; so the voltage V- at the inverting input and the output voltage VOUT are zero as well.
VIN = 16 V (at the first moment): Then imagine the input voltage "jumps" up to 16 V. The "lazy" amplifier does not immediately react and, at the first moment, its output voltage remains zero. The potentiometer acts as a voltage divider with ratio 0.5 driven from the left and the voltage at the inverting input sharply "jumps" up to 8 V.
Transition (shortly after): After a while, the amplifier "recovers" and begins acting to set its output voltage VOUT = -K.V- = -6.8 = -48 V. It begins to lower VOUT with the highest speed possible. The potentiometer acts again as a voltage divider with ratio 0.5 but now it is driven from the right. So the voltage at the inverting input proportionally follows VOUT and also goes down.
Now the most important for understanding: During the transition the amplifier is not an amplifier; it is rather an integrator. It does not manage to set its output voltage VOUT = -6.V- as it should... VOUT is less than needed. This is not a stable state... and there is no equilibrium. But VOUT is "moving" toward the equilibrium and the op-amp strives for the point of equilibrium. The transition will continue during VOUT/V- < K.
Equilibrium. When VOUT/V- = K, the op-amp output voltage will stop changing and equilibrium will be established (2 V at the input and -12 V at the output).
So, during the equilibrium, the amplifier ceases to be an "integrator"; it becomes again an amplifier. This is the moment "when the cycle of stabilizing ends"... and this is the answer of the question.
The voltage diagram illustrates geometrically this relation by two similar right triangles with legs V- and VOUT.
Varying K: We have used a small amp gain of 6 to see the voltage V- at its input. But this voltage is undesired; we want to zero it (virtual ground) so that the overall gain of the circuit will be exactly R2/R1 (the idea of the inverting amplifier). The only way to (almost) zero it is by increasing the amp gain K. So let's begin increasing K...
The amp will further lower its output voltage to keep the proportion of 6 between V- and VOUT. As a result, V- will further decrease... and when K becomes (almost) infinite, V- will be (almost) zero. Then the proportion between the two voltages will be the well-known VOUT/VIN = - R2/R1... and there is an equilibrium as before... and a virtual ground at the amp input. Now we can replace the humble amp by the more sophisticated op-amp. It has an additional non-inverting input... but we do not need it... so we simply connect it to ground.
The voltage diagram illustrates geometrically this relation again by two similar right triangles with legs VIN,R1 and VOUT,R2.
After four years...
I am returning to this interesting experiment to carry it out now using the CircuitLab simulator. It is much more convenient for this purpose than the thought and real experiments above.
Implementation
Quantities: The values of quantities were a little strange in this geometric interpretation because they had to correspond to a certain number of boxes on the squared paper. Here this consideration is dropped and I have chosen more convenient (multiples of 10) values.
Amplifier: I have used an inverting amplifier with a low (K =10) fixed gain. Since there are not such an element in the CircuitLab library, I have decreased the huge gain of an op amp to 10, and grounded the unnecessary non-inverting input.
"Amp man": But the most exciting experiment I propose to do first is we to play the role of the amplifier. This way, being the "lazy human being who thinks slowly", we will best understand what the op-amp is actually doing to reach equilibrium.
Conceptual circuit
To do this, we can produce the output voltage by a variable voltage source Vout aiming to keep it 10 times greater than the midpoint voltage V- between the resistors R1 and R2. We can distinguish three typical situations during the transition:
t1: Vout/Vin = 0. At the first moment, Vin "jumps" to 1 V but we fail to react and Vout = 0 V. V- becomes 500 mV (half of Vin "produced" by the voltage divider R1-R2).
simulate this circuit – Schematic created using CircuitLab
t2: Vout/Vin = -5. After a while we recover and looking at the positive reading of the voltmeter begin to drop Vout in the negative direction with the goal to reach the "gain" of Vout/V- = -10. Unfortunately, V- also begins to descend, albeit more lazily, and a kind of "vicious circle" results. As a result, at this moment, we have only managed to achieve gain of K = -714.3/142.8 = -5.
simulate this circuit
t3: Vout/Vin = -10. We do not forget that we are a "10x-inverting amplifier", and keep changing Vout in the negative direction. And since we're too lazy to adjust or calculate Vout, we can just "steal" it from the next schematic (use the op-amp as the "analog computer" that calculated it). Anyway, finally our efforts are crowned with success and we reach the desired ratio of K = -833.3/83.33 = -10.
simulate this circuit
Hmmm... we came to an interesting conclusion - the role of an (op-)amp in a negative feedback amplifier is to keep the Vout/V- ratio equal to its open-loop gain. Only then is it in equilibrium and all relations between quantities apply. This definition is much more precise than "the role of an op-amp is to maintain zero voltage (virtual ground) at its input". We will have to apply for this wisdom to be written down in thick textbooks to go down in history -:)
Op-amp circuit
Now all that is left is to "hire" , for minimal pay, some amp to do (quickly, accurately, and without grumbling) this routine.
K = -10: At the minimum gain of 10, the input error (V-) is only 83 mV.
simulate this circuit
K = -1000: If we increase the gain 100 times, V- becomes only 1 mV.
simulate this circuit
K = -100000: If we finally put an op-amp with the minimum for it gain of 100000, V- becomes negligible (10 μV).
simulate this circuit
Let's investigate the role of the gain by sweeping it from 1 to 1000.
Where is the true virtual ground?
The significant input voltage V- gives the impression that there is no real (0 V) virtual ground. Actually there is, but it has gone inside R2. To "see" where it is, we should "open" R2. However, there is also an indirect way - by connecting a "measuring potentiometer" P in parallel to the circuit (potentiometer) R1-R2. Then we can "move" its wiper (change its K) to find the virtual ground point. For example, for the schematic below, you have to set K = 0.545454562 to see this zero voltage point.
simulate this circuit
Also, we can sweep the P's K (wiper position) to see the voltage distribution along the R1-R2 potentiometer and where it crosses the zero-voltage level (virtual ground point).