Re: 75 grain A-Max in 9" twist
Ok. Good answer, but totally wrong.
When a bullet emerges from the bore, it is at maximum velocity both in forward flight and rotation. The spin rate of the bullet is a function of the muzzle velocity and twist rate; for purposes of illustration, lets take a bullet in a 1:12(*) twist going at 3000 FPS. Its spin rate is 3000 RPS (revolutions per second.)
The reason a bullet needs to spin is to keep it point on during its flight, using the gyroscopic effect. Without this, the air resistance the bullet encounters would try to destabilize it and make it tumble, because the bullet is not a sphere. Longer, more aerodynamic bullets are more unstable and require a faster spin to stabilize. Of course, the faster the bullet goes, the greater the air resistance and thus the greater the tendency to tumble so a faster spin rate is required for that.
Now, as the bullet we talked about emerges from the barrel, it is going forward at 3000 FPS and spinning around its axis a 3000 revolutions per second. That is the fastest it will ever go as the air resistance is now acting upon it at full strength and slowing it down in its forward motion. However, there is really nothing to slow down its rotation, so it continues to spin at 3000 RPS. But as its forward velocity drops rapidly (and that's where the BC of the bullet comes in) the tendency for the bullet to tumble also drops. Since the spin rate remains the same but the bullets is slowing down, the bullet becomes more stable as time (distance) grows.
Provided there is no event to cause the bullet to tumble, such as going transonic for certain bullets, it is safe to say that if a bullet is statically stable at the muzzle, it will be staticaly stable throughout its entire flight. (Please don't bother with the shooting from the side of a cliff or a mountain, or an aircraft.)
(*) Fixed typo pointed out by FCS below.