First of all, observe that multiplication of coordinates preserves equality of distances: if points $a$ and $b$ are equidistant, then so are $\alpha a$ and $\alpha b$ for any number $\alpha$. Thus, we can reduce the problem to the case when the radius of the circle is equal to one.
Now, consider a function $f(\cos \varphi, \sin \varphi) = (\cos(n\varphi), \sin(n\varphi))$. As you can see, it wraps the circle around itself $n$ times, and $\left|\frac{\mathrm{d}}{\mathrm{d}\varphi}f\right| = \left|(-n\sin(n\varphi), n\cos(n\varphi))\right| = \sqrt{n^2\sin(n\varphi)^2 + n^2\cos(n\varphi)^2} = n \sqrt{\sin(n\varphi)^2 + \cos(n\varphi)^2} = n,$ so as $f$ wraps a circle around itself, it does so with constant velocity, so each 'lap' has the same length. Thus, for example, all points that $f$ maps to $(\cos \varphi_0, \sin \varphi_0)$ must be equidistant, and there are exactly $n$ of them.
So now all that's left for you to do is to solve the equation $f(\cos \varphi, \sin \varphi) = (\cos \varphi_0, \sin \varphi_0)$, and you'll arrive to Berci's answer.
One last thing, if you're familiar with complex numbers, $f(z) = z^n$, so any set of $n$ equidistant points on a unit circle is a set of $n$-th roots of some number of length one.