Thinking aloud, (lets pretend op amps can deal with amps of current- it makes the maths easier). If you use the common terminal of a +- supply as one terminal of the resistor. Then the other terminal goes to the op amps output via a resistor. the actual input point now monitors the voltage. So if its set to 1milli ohm, if 1mV is applied to the input, the opamp output must "****" 1A out of the applied voltage source. You can see where this is going. . if 2mV is applied, the amp must **** out 2A. . . Now if we reduce the circuits gain so that 2mV "sucks" out 1A, it will look like 2 milli ohms.
So now design the circuit, I think you will have to use instrumentation amplifiers, as 10 mA X 1 milli ohm = 10 micro volts. So suppression of noise is paramount.
Frank
**** = s u c k