We ran 6-#1 Copper THHN's to the motor today, and I have a clear mind about it because I now know that 3 of the "half size" wires (#1) will be adequate to carry the less than FLA start up current while it is "wye", and that parallel runs of #1 will carry the 180 Amp load when it is on "run" and in the delta configuration.
A small clarification: None of the 6 leads are actually electrically in parallel, meaning connected at both ends to form a single conductor. Instead each lead is connected to a separate coil terminal. In a standard delta connected motor with three leads coming out, each lead is connected to a _pair_ of coil ends (well really sets of coils, each comprising one phase), and each lead carries the vector sum of the current from the two coils. Since the two coils are out of phase, the net current from the pair is 1.732 times the current of either one.
Said differently, you might have 10A flowing though a coil set from phase A to phase B, 10A flowing from phase A to phase C, but 17.32A flowing in the phase A lead to the motor.
But in a wye start/delta run motor, each wire gets directly connected to one coil, and the currents get joined in the contactor. Each conductor has to carry the full current of the respective coil. In the example above, each conductor has to carry the full 10A, not half of 17.32A. This is why you have the factor of 0.58, rather than a factor of 0.5, and also why you can 'parallel' conductors smaller than 1/0 in this case. (1/1.732= 0.577)