Transistor control through opamp
This circuit consists of two cascaded dual circuits - current-to-voltage converter (Rs) and voltage-to-current converter (the op-amp, Q1 and R3). So it is a current-to-current converter... current amplifier... current attenuator or, if the output current is equal to the input one, the so-called current mirror.

Simply speaking, the input current creates a voltage drop Vs = IIN.Rs across Rs and the op-amp follower copy this voltage across R3 by the help of the output (collector) current IOUT. So IOUT.R3 = IIN.Rs -> IOUT = IN.Rs/R3 = IIN/2000, i.e., the circuit is a current attenuator.

The clever trick of this circuit solution is that the emitter follower (Q1) is put in the feedback loop and the op-amp compensates the transistor base-emitter voltage VBE. For this purpose, the op amp "lifts" its output voltage with VBE; so the emitter voltage is equal to the input voltage (across Rs). Thus the combination of the op-amp and emitter-follower can be considered as a perfect voltage follower working on the resistive load R3... i.e., perfect voltage-to-current converter (current sink).

It is interesting to see that both resistors Rs and R3 act as current-to-voltage converters (V = I.R) but the function of R3 is "reversed" by the negative feedback... and it acts as the dual voltage-to-current converter (I = V/R). This is a fundamental property of negative feedback systems. The op-amp non-inverting amplifier is another example of this unique feature where the attenuating voltage divider is made to act as an "amplifying voltage divider".

The resistors R1 and R2 do not play any role since the op-amp inputs have high resistance and should be removed.