Below is an EVL8030 for MP8030 POE PD and DC/DC controller. They have put a MOSFET on the secondary winding. At first I thought it was used as a snubber but I'm unsure about this. It's a flyback buck current switch mode power supply.
Some extra details: T1 = Lp = 55 μH, turns ratio Np:Ns1:Na:Ns2 = 18:6:5:5, 12 V output at 6 A.
It’s an active rectifier. For low output voltage, the forward voltage drop of a regular diode translate to a significant hit in efficiency.
Let me give you an example: A Schottky diode have about 0.4 V forward voltage + some series resistance. Let's take the SDT10100P5 at 6 A and 85 degrees, it will have about 530 mV of forward voltage, translating to about 3.2 W of dissipation.
Going for a MOSFET instead, the BSC070N10N has 7 mΩ rated resistance. Running somewhat hot, it will however increase to 9.5 mΩ. At 6 A, this translates to about 340 mW of dissipation.
The active rectifier have almost one tenth of the losses of the Schottky, so good savings on thermals. It is however more expensive than the diode and needs additional circuitry from the extra secondary winding and parts between that and the MOSFET gate.
What's the purpose of the MOSFET on the output of the flyback buck converter in this circuit?
It's acting as a synchronous rectifier. See the body diode I've highlighted in blue: -
By itself, the blue diode is how you would place a regular diode in the circuit. The MOSFET replaces the regular diode and, when expected to conduct, the MOSFET shorts out its own body diode giving lower conduction losses than a regular diode.
