homework and exercises - Does a clock oscillating in a friction-free hole through the center of a planet run slower than a stationary clock on the surface?

Assume a clock is dropped into a friction-free hole through the center of a symmetric, non-rotating planet, far from any other massive object. Clearly, the clock oscillates from one end of the hole to the other forever, and, as it is subject to time dilation due to relative speed and acceleration, it runs more slowly than a stationary clock at the center of the planet. But, does it run more slowly than a stationary clock at the surface, which is subject only to the acceleration of gravity, or does it keep exact time with the surface clock? I think it runs exactly the same as the surface clock. Am I correct?

[Added 10/20/2015 4PM] Thanks for the initial answers, but they indicate I need to clarify my question and explain my reasoning.

1) My "planet" (unlike the Earth) is "symmetric" (perfect sphere of constant density), "non-rotating" and "far from any other massive object" (not in orbit around a star).

2) At the instant the clock is dropped into the friction-free hole, it is adjacent to the surface clock and they are both set to zero. When it first passes the center clock, that clock is set to whatever time is on the oscillating clock.

3) After many cycles, as the oscillating clock passes each of the other clocks, the times are compared and, if the oscillating clock time is substantially lower, it is said to have been subject to more time-dilation than the center and/or surface clock, and vice-versa. If the times are equal, or nearly so considering the accuracy and precision of the clocks, it is said to have the same time-dilation. I believe the oscillating clock will have run more slowly than the center clock (more time-dilation) and equal to the surface clock, for the reasons in the following points. (Please correct me if I am wrong.)

4) Let the origin of the reference inertial frame be the center of the planet. The center clock is "at rest" in that frame, at zero speed and zero gravity. The surface clock is not moving, but it is subject to gravity, and therefore more time-dilation than the center clock.

5) The oscillating clock is subject to varying amounts of speed and/or gravity. When adjacent to the surface clock, it stops momentarily and reverses direction. At that instant, it is at zero speed and the same gravity as the surface clock, and so running at the same rate (same time-dilation) as the surface clock.

6) When the oscillating clock is adjacent to the center clock, it is at maximum speed and zero gravity. What is that maximum speed? By conservation of energy, the kinetic energy of the clock, as it zips past the center, is exactly the same as the gravitational potential energy the clock had when dropped from the surface. Therefore, the maximum speed just happens to be the escape velocity at the surface, since escape velocity perpendicular to the surface is defined as the speed where kinetic energy is exactly equal in magnitude to gravitational potential energy, and (assuming no atmosphere) a rocket will continue its journey into space forever.

7) Time-dilation due to gravitational potential energy of a stationary clock on the surface of a planet is exactly equal to the time-dilation due to kinetic energy of a rocket in deep space moving at the escape velocity corresponding to the surface of that planet. By conservation of energy, as the oscillating clock shuttles back and forth between the ends of the friction-free hole, its total energy (combination of kinetic and gravitational potential energy) remains constant, and exactly equal to the gravitational potential energy of the surface clock. Therefore, I believe the oscillating clock will experience exactly the same time-dilation as the surface clock.