Why starting current in star-delta starter becomes 1/√3 times that of DOL starter?
The whole idea is that a smaller voltage is delivered to the motor windings during start up.

The end game is that the motor (or load) is configured as a delta winding formation. In other words, after the load is up and running, each winding receives line voltage.

However, during start-up the windings are put into star formation and each winding will receive 57.7% of the AC voltage.



When starting the motor (above) KM3 and KM1 activate whilst KM2 is deactivated. You should be able to see that KM1 terminates the motor windings into a star point thus the supply voltage to each winding is only 57.7%.

After a short while KM1 deactivates and KM2 activates. This puts all three motor windings into a delta configuration thus each winding receives 100% line voltage.

Here's a picture that shows (or implies) that math: -



Here's the SE question where I borrowed the final picture from.