On U146, pins 3 and 4 are high voltage power in,
Correct, rectified hi V from mains.
... then after a winding pin2 goes to the MOSFET drain.
Correct. So if pin 2 and pin 3 or 4 is one windings, then its on/off will be switched by the MOSFET, used to generate induced voltage on the secondaries.
And there's an auxiliary winding between p5 and p6 on GND. Some of the energy that gets stored in the main winding, get transfered into this little winding, and then typically around 10-15V is created and rectified by D115, and smoothed by 100uF/25V near by.
This is where the difficulties come from. I've had a closer look at the poles of the transformer trying to deduce the internal wiring.
I use numbers to indicate the sizes of the wires connected to the pins, the smaller the number, the thicker the wire. And I use "2x" to indicate there is two wires to one pin (of same size)
On the primary side:
Pin 2: 1
Pin 3: 1
Pin 4: 3 (2x)
Pin 5: 2
Pin 6: 2
So pin 2 and 3 has to be of one winding, so are pin 5 and 6. What's puzzling is that how can pin3 be connected to both ends of the same winding?!
On the secondary side:
Pin 7: 1
Pin 8: 1
Pin 9: 2, 3
Pin 10: 2
Pin 11: 1, 2(? not very sure)
Pin 12: 1
This is more difficult: One possibility is that pin3 7/8 are one winding, so are pins 11/23, and pins 9/10. But again, where is the single size 3 wire for?! (Unless the size 2 wire on pin 11 is actually size 3, going to pin 9.)
There is no way that one end of the winding of the primary side goes to the secondary side, that will defeat the isolation. And I checked with DMM. Absolutely both sides are isolated.
It looks like this weird thing – attached.