Do extended bodies in a gravity field fall slower or faster than point-like structures? The weak equivalence principle states that all point-like particles fall along the same in a gravity field.
If a body is extended it experiences a tidal force which causes the bodie to elongate. If this elongation is countered by ìnternal forces, parts of the body will fall differently from a freely falling particle.
If we compare the center of mass of the extended body with a free point-mass (having the same mass as the extended body) when they fall in a strong gravity field (say close but outside a black hole), will they fall identically? For example, when they start falling at the same height, will they arrive at different heights after a fixed time? Will the motion of the COM of the extended body be affected by the motion of the other parts?
 A: Yes, extended objects can follow different trajectories in general relativity due to different shapes or rotations.
Perhaps the most well-known case is that rotating objects near a Kerr black hole follow different trajectories (that can be chaotic).
One can also derive formulas for how the centre of mass moves for small objects (eq 2.23), showing deviations from the expected trajectory due to the quadrupole moment of the mass distribution. Indeed, near a Schwarzschild black hole and without spinning, it is possible to change the "effective mass" determining the trajectory by changing quadrupole moments (indeed, the authors note that this is akin to the changed potential energy due to being an extended object in Newtonian gravity). This can speed up or slow down radial falling (eq. 4.17).
A: If rotation is allowed, they fall slower. During the fall, the conservation of energy reads:
$$   \frac{1}{2} m.V_{cm}^{2}+\frac{1}{2} J. \Omega ^{2}+ m~g~z_{cm} =E_{0}$$
Where:
$\frac{1}{2} m.V_{cm}^{2}$ is the kinetic energy of the center of mass.
$ \frac{1}{2} J. \Omega ^{2}$ is the rotational kinetic energy.
$  m~g~z $ is the gravitational potential energy.
$ E_{0}$ is the initial mechanical energy.
So:
$$V_{cm}(t)  =\sqrt{ \frac{2}{m}  \big(E_{0}-m~g~z_{cm}(t)-\frac{1}{2} J. \Omega ^{2}(t)\big) }$$
