What happens to a particle in the exact center of a Kerr black hole? Kerr black holes (and Kerr-Newman black holes), instead of the "point" singularity theorized in spherically symmetric black holes, instead have a "ring" singularity, spread along the equatorial plane of rotation. any particle inside a spherically symmetric black hole will tend to go towards the center. But what if a particle is in the exact center of a Kerr black hole? Will it go towards any point in the ring singularity, or will it stay where it is, unless it is perturbed in one direction?
 A: A maximally extended Kerr solution has multiple horizon. The outer horizon is an event horizon and the inner horizon is a Cauchy horizon.
Outside both is a normal type region of spacetime. In between the two the r coordinate is a timelike coordinate so you must continue decreasing r if you entered with a decreasing r.
But inside the Cauchy horizon means the singularity is on your past light cone. So its not really sensible to say you know what would happen.
Another indicator of a problem is that on the other side of the ring singularity is a region with closed timelike curves. So you can see a singularity (what does that look like) and you can see a region beyond it with time travel (what does that look like). But you can still have a manifold and a metric, this is after all a solution for an external black hole so it just was always there, spinning around.
If you look at the center of the ring you can have a geodesic that stays at the center. But is that really meaningful?
A: The particle would be in an unstable equilibrium (there would be no prefered direction if you are at rest), but if it were slightly off center it accelerates outwards (gravity is repulsive there, see the d²r/dτ² in the numeric display). If I start at r=0, θ=0 the particle stays where it is, but if I displace it slightly (r=1/1000GM/c², θ=1/1000rad, v=0) it accelerates outwards. 
The escape velocity (to escape the time reversed white hole, see the last paragraph) in the center goes to 0, so a very small displacement is already enough to shoot the particle away. The coordinates used are Kerr Schild (click on the image to enlarge):

The frame dragging rotation in the center also vanishes, so if the test particle is small enough it may stay in the exact center without spinning, see ω fdrag and v fdrag (angular and true frame dragging velocity) in the numeric display (the local velocity components on the bottom right are relative to local ZAMOs, which are possible inside the inner ergosphere).
The acceleration is always in the positive r direction, since when you pass through the ring into negative space the sign of gravity is reversed. So if the displacement is in direction of the negative space you still get accelerated in the direction of positive space. 
Then the particle goes back up again until it reaches the inner side of the Cauchy horizon, where its proper time goes to zero while the Finkelstein time continues, which means that the particle gets hit by all the light that falls into the black hole at once, so he gets fried due to the infinite blueshift. 
If you somehow survive the infinite blueshift on the Cauchy horizon your proper time gets reversed relative to the universe after you saw all its future in one moment, and you can escape the black hole again, which is now a white hole in your frame of reference. But that happens far away from the center, where everything should be fine for the observer.
A: The short answer is, what happens is pretty much the same as if the particle were at the exact center of a Newtonian black hole or planet instead: experiencing maximal pressure and no net gravitational attraction, it would remain in stable equilibrium.
The long answer adds, you describe a Kerr black hole as I would a Kerr vacuum. To make the long answer short, there is no such thing as "the exact center of a Kerr vacuum" so what happens to a particle there is irrelevant.
In extenso, the Kerr vacuum is only the exterior part of the black hole model, the part where there is no gravitating matter and where the RHS of the EFE is 0. The gravitational field must have sources though, so the model must be complemented with an interior solution (it is interior because the S-E tensor is non-zero there, as in the interior of a planet) and with a suitable continuity condition at the boundary of the two domains.
The distinction is not only for the sake of pedantry. The ring singularity is not physicist-friendly, so interior solutions are often required to be regular in a domain enclosing the singular ring; in much the same way an interior solution to Poisson's equation is regular in the domain where it replaces Green's singular solution. Once the ring singularity is thus quietly removed, it is pointless to ask whether a test particle can fall toward it.
Interior solutions confined to the space below the singular ring, sometimes called the negative space because it is where the $r$ component of the B-L coordinates is, well, negative, have a number of other embarrassing properties than they leave the singularity naked. One of these is, there is no such thing as "the" center of symmetry of the exterior solution, meaning "the" center of the singular ring or, in B-L coordinates, the center of "the" disk $r=0$.  Two distinct points have equal claims to this exalted position: one at $r=0, \cos \theta = 1$ and one at $r=0, \cos \theta = -1$; none of them is "the" center of the interior solution (supposedly confined to $r < r_0$ for some $r_0 < 0$). If you drop a particle in free fall from rest at any of these positions, it will enter negative space at once and crash onto the boundary of the interior solution, not run toward the singular ring.
What you call "the center of symmetry of the black hole", if it exists, can only be the center of symmetry of the interior solution; whereas your description suggests the vacuum solution has a center.
