No, the resistance of the shunt built inside the DMM is a few mΩ, while the external resistor across which you are measuring the voltage drop is 5Ω.
The final current in the series circuit when the circuit is pulsed is determined by the inductance, the length of the pulse, and the total resistance in the loop. In one case the resistance is the DC series resistance of the inductor plus a few mΩ, while in the other case it is the inductor resistance in series with 5Ω! Why would the current not be different?
btw-to measure the current in a circuit without changing the current by inserting a current shunt resistor, the shunt resistance must be much less than the circuit resistance. In your case the DC resistance of the inductor is probably a couple of Ω. Try making your current shunt resistor 10mΩ or 100mΩ.