Does the center of a star age faster than its relativistic equatorial surface? If so, how does it maintain its speed and spherical shape? Imagine a massive (the diameter is millions of miles across) star rotating on its axis at relativistic speeds. Assume, now, that the center of the star is stationary. Note that the center of the star ages faster than a point in the peripheral equator of the star, as the center is at rest. I assume that time for the center passes faster than at the equator. I expect that the equator will change its shape to be narrower and never achieve a spherical shape. If the above thought is correct, how would such a star survive? Would the decay of the mass in the center occur faster, collapse faster or split in half? And if so, how is the center of the sun's gravity stronger than at its surface? Remember that we are talking about a star spinning at relativistic speeds.
 A: 
Quote: "Imagine a massive (the diameter is millions of miles across) star rotating on its axis at relativistic speeds. Assume, now, that the center of the star is stationary. Note that the center of the star ages faster than a point in the peripheral equator of the star, as the center is at rest. I assume that time for the center passes faster than at the equator."

That is correct. If we neglect gravitational time dilation (which, since you tagged your question with "special relativity" instead of "general relativity" we seem to do) and focus on the kinematic component the clocks on the equator run slower by a factor of
$$\frac{\tau}{t} = \sqrt{1-\left(\frac{\omega \ r}{c}\right)^2} = \sqrt{1-\left(\frac{v}{c}\right)^2}$$
where $\omega$ is the angular and $v$ the tangential velocity. If $\omega$ and $v$ are constant you can multiply it out, if the angular velocity and/or the radius changes over time you have to integrate over time
$$\frac{{\rm d}\tau}{{\rm d}t} = \sqrt{1-\left(\frac{\omega(t) \ r(t)}{c}\right)^2} = \sqrt{1-\left(\frac{v(t)}{c}\right)^2}$$
Since the clock on the equator always comes back to the point where a stationary clock would show the same time as the observer in the middle's clock, we have a twin paradox, where the travelling twin who returns to his starting point (the one on the equator) is the younger one, while the twin who stays at home (the stationary one) is the older one.
So the equation we know from special relativity also holds in this scenario, with the difference that the time dilation between our two clocks is not relative, but absolute.

Quote: "Would the decay of the mass in the center occur faster, collapse faster or split in half? And if so, how is the center of the sun's gravity stronger than at its surface? Remember that we are talking about a star spinning at relativistic speeds."

If we are talking about a fast spinning star where we can not neglect the gravitational component of the time dilation any longer, the gravitational acceleration in the center is 0, but the gravitational potential is highest (and the potential is responsible for the gravitational time dilation).
For a rotating black hole we can calculate the gravitational time dilation of a frame dragged shell observer relative to the external observer, who is stationary relative to the fixed stars (the coordinate bookkeeper), which is in natural units of $G=M=c=1$:
$$\frac{{\rm d}\tau}{{\rm d}t} = \sqrt{\frac{(a^2+(r-2) \ r) (a^2 \cos ^2(\theta )+r^2)}{{(a^2+r^2)^2-a^2 (a^2+(r-2) \ r) \sin ^2(\theta )}}}$$
which in the nonspinning case where the spin parameter $a=0$ reduces to
$$\frac{{\rm d}\tau}{{\rm d}t} = \sqrt{1-\frac{2}{r}}$$
where it already takes an infinite amount of coordinate time $t$ time until anything hits the horizon, since the time dilation there goes to infinity. If the shell observer also has a local velocity relative to the frame dragged shell that corresponds to his radial distance, you divide the equation above by the Lorentz-factor.
For the inside we only have a solution for the nonrotating Schwarzschild case, while for the rotating Kerr case we only have a vacuum solution from the outside, so we can describe how fast the time runs for an observer on the surface of such a collapsar. Mathematically, after a test particle crossed the horizon (which takes a finite proper, but an infinite amount of coordinate time), the coordinate time runs backwards again while the proper time increases. But this is just an artefact which does not really make any physical sense, so we can only thrust the exterior-solution in the strong relativistic and fast rotating case.
It's hard to say if the gravitational time dilation in the center of a collapsar is stronger, or weaker than the combined effect on the surface, since the rotation is higher on the surface, but the potential is higher in the center. Since I don't know a valid solution for the fluids inside the Kerr-interior I can not really answear part 2 of your question.
