Switching 12V with an active low 5V signal

How about this for an idea.

With the input at 5V no current can flow through the zener (5 + 9 > 12). The output PNP transistor of held OFF by the base emitter resistors (= 4k7 + 2k2) and the output is 0. When the input is pulled down to 0V a small current will flow through the base and the 2k2 resistor. The junction of the two resistors will be a 9V (the zener voltage) and the base will be at 11.4V (assuming a 0.6V Vbe drop). The total current flowing through the base and zener (sink current) will be added (Kirchoff's current law). With values shown the base current will be 0.5mA and the resistor current 1.4mA giving a sink current of just under 2mA.

You can achieve this with two NPN transistors as shown below. As you can see, whenever a 0V signal is given, 12V is seen in the output, and whenever a 5V is given in the input, 0V is seen on the output.

Let's see how it works. First of all, let's start with the scenario where the input is 5V, or in other words, HIGH. This will turn Q1 ON and the voltage on the collector of Q1 will be almost equal to the voltage on its emitter, which is GND. Q2's base is connected to Q1's collector, so when there is 0V on Q1's collector, or in other words when Q1 is ON, Q2 is OFF. That is because Q2's base will be shorted to ground.

When the input is 0V, or LOW, Q1 will not turn ON and can be imagined as being not connected at all. So, the current passing through R1 will turn ON Q2.

The current of Q2 is limitied with its base current and hFE, as it can bee seen in the below equation;

\$I_{CQ2}=I_{BQ2}*hFE_{Q2} = \dfrac{12-0.6}{10*10^3}*300=350 mA \$

As it can be seen, maximum current that can pass through Q2 is about 350mA. But this highly depends on hFE of the transistor, which can vary anywhere from about 50 to 300. With a hFE of 50, the current can be maximum of about 60 mA, which is enough for your specs. Lowering R1 will increase the current that Q2 will pass.