What do black holes spin relative to?
But if you start running around it, it will move faster or slower relative to you. In this case, the disc has a ground speed, 60 rpm, because it has something to spin in relation to, in this case, the table.
Actually, this is fundamentally incorrect. The spinning of the disk has nothing to do with the table in principle. Acceleration, including spinning, is not relative. It can be measured without reference to any external object. For example, using a ring interferometer, or a gyroscope.
It does not matter if the object is a disk or a black hole or anything else, spinning is not relative like inertial motion is.
When I move around the black hole, the black hole spins slower relative to me, and consequently has a larger event horizon.
The event horizon is a global and invariant feature of the spacetime. Your motion does not change it. Of course, you can use whatever coordinates you like and make the coordinate size change as you wish. However, which events are on the event horizon is unchanged by your motion.
This is just Newton's bucket in modern garb. The best explanation of this effect that I have seen is in Carlo Rovelli's book Quantum Gravity, which explains it as rotation with respect to the gravitational field. According to Einstein's Theory of General Relativity, the gravitational field is a real physical entity. And Rovelli says about Newton's bucket (on page 56 of the 2005 hardback edition):
Einstein's answer is simple and fulgurating:
The water rotates with respect to a local physical entity: the gravitational field.
Rovelli regards this as so important that he underlines it, as well as putting it in italics; but my formatting skills don't run to that. And yes, fulgurating is a real word.
What do black holes spin relative to?
Relative to an inertial reference frame infinitely far from the hole, in which the hole has no translational motion.
And what happens if you move around it?
A spinning black hole is azimuthally symmetric. It “looks” the same from any azimuthal angle. Its spin parameter $a$ in the Kerr metric has nothing to do with how fast you move around it.