Dc-Dc converter from 12v Input 400v Output
Flyback transformers are best viewed as two inductors, wound on the same core.  The primary winding charges up the core, and the secondary dumps the charge into the load.  Using the data sheet on your link, we can go through some rough calculations for this transformer.



The winding ratio is 135:9, and inductance is proportional to the square of the turns for a given core.  Since these are wound on the same core, the ratio of inductance is 18225:81.  Since the 135-turn coil has an inductance of 2100 uH, the 9-turn coil on the same core can be approximated at 9.3 uH.

Your load is 450 volts at 0.1 amps, or 4.5 watts.  One watt is one joule/second, and this transformer appears to be happy at 132Khz, so neglecting losses, each time the transformer switches, it must pass 4.5 watts/132,000 or 34 microjoules to the load.

Since the energy stored in the 9.3-uH coil is (1/2)Li^2, this means that the 9.3-uH coil must reach a peak current of 2.7 amps each time it switches in order to have the 34 microjoules to dump into the load.

So, the transformer coil, rated at 1.5A, will likely either heat up or saturate.  You can apply this approach to other transformers you find. You can see that you will have to add switching losses and the diode drop effect, but this will get you in the ball park.

Good Luck!