How can I switch on 12V 500mA supply to a load when a signal is taken from 5V down to ground?
That won't work. The pin would have to go up to +12 to turn the load fully off, which would more than likely fry whatever is in the black box.
You can do something like this:



simulate this circuit &ndash; Schematic created using CircuitLab
Pick a MOSFET with low enough Rds(on) to handle your load with very little dissipation, and that can withstand Vgs of something like +/-20V. If its Rds(on) is specified at 10V that will allow a bit of drop in the 12V bus.
How it works: When the input is high (+5V) Q2 is turned on via base current through R3 (about 430uA). That pulls the base of Q1 down to 100mV or less, sinking the 1.3mA through R2. Q1 then turns off allowing the gate of M1 to be pulled up to +12V through R1 and M1 turns off, turning the load off. The opposite happens when the input is low, Q1 is on and the gate of M1 is pulled down to near ground, so Vsg is almost 12V.