Why don't objects following a geodesic maintain their rotational state? If I throw a ball into the air, it comes back down because that is the shape of spacetime and the ball is just following it.
But if I paint a spot on the ball and throw it upwards with no rotational motion, even when the ball comes back down the position of the spot stays the same relative to me.
If the ball is following a continuous path along a geodesic I would expect that the rotational state of the ball stays constant relative to the geodesic, in other words, if the spot faces up when the ball is launched it should face down when the ball comes down.
The observed behaviour would suggest that there are forces in play to rotate the ball to keep its original rotational state.
Are there other forces acting on the ball I'm not aware of?
 A: You're considering this in the noninertial frame of the earth, which makes it more confusing. In GR, we consider free-falling frames of reference to be the inertial frames. In such a frame, the ball's center of mass follows an inertial path, which looks like a straight line in spacetime. Meanwhile, the spot painted on the ball follows a parallel path alongside it. There is no mystery, and everything is as we expect.
If you like, you can define a spacetime displacement vector $\Delta \textbf{x}$ from the ball's center of mass to the spot, at times that are simultaneous in the ball's inertial frame. This four-vector is parallel transported, and again that makes sense. In this parallel-transport process,  $\Delta \textbf{x}$ stays the same, so the ball maintains the same orientation relative to the distant stars.
Now let's switch back to the accelerated frame tied to the earth's surface. In this frame, $\Delta \textbf{x}$ has a nonvanishing time component. The place where your intuition might be leading you wrong is that you might be expecting that in the earth's frame, the spatial part of  $\Delta \textbf{x}$ would still maintain the same angle with respect to the spatial part of the ball's velocity vector. This is not the case. When you switch frames, what stays invariant is the inner product of the four-vector $\Delta \textbf{x}$ with the four-velocity. If that is to happen, then it's not possible for the inner product of the spatial parts to stay the same.
A similar example would be the orientation of the gyroscopes aboard the Gravity Probe B satellite. Their spatial orientation maintained approximately the same direction relative to the fixed stars, not relative to the satellite's spatial orbit.
A: Which rotational state? When the ball goes up, then falls, it obey the energy conservation. It goes up to a point where all its energy becomes potential, then returns and its potential energy transforms into kinetic. From the beginning the ball has no rotation movement, its angular velocity is zero, and so the energy of rotation.
There is complete disconnection between the center-of-mass movement of the ball, and its dynamics IN the center-of-mass. Inside the center-of-mass it is NOT KNOWN that the ball goes up or returns.
