Kerr Metric from rotated Schwarzschild?

One of the key features of the Kerr metric is that the black hole horizon is rotating with respect to the space at infinity -- you get frame dragging effects that cause the notion of "rest" to change as you get closer to the black hole. In fact, energy must be exerted if you want to stay stationary with respect to infinity, until you finally reach a surface, called the ergosphere (which is actually outside the event horizon), where it is actually impossible to be at rest relative to infinity.

These are all physical, frame-independent effects. In particular, it is possible to transfer energy from the black hole to infinity by clever explosions within the ergosphere, through soemthing called the Penrose process. A mere coordinate transformation would not be able to replicate effects like this. And a coordinate change describing a simple rigid rotation around the Schwarzschild black hole would leave "the sphere at infinity" rotating at the same rate as the hole.

I think the real question is, why would it be? A rotating coordinate frame is not the same as physically rotating object. This is easier to see in Galilean relativity, where we know perfectly well that only uniform motion is relative: a rotating star is not the same as an observer rotating around a static star, because the latter experiences a centrifugal force.

Suppose we take the Kerr and the rotating Schwarzschild metrics, which according to you should be the same, and let the black hole's mass go to zero. The Kerr metric goes to the Minkowski metric, which is reasonable, since you're standing still in empty spacetime. But the rotating Schwarzschild metric goes to a rotating Minkowski metric, which is different from regular Minkowski! You have centrifugal forces and so on. Therefore, the original two metrics are not the same.

Moving to a rotating coordinate system is not a symmetry of nature. That's all there is to it, really.