If $$S = \{z \in \mathbb{C} : |z - i| = |z + i| = |z - 1|\}$$, then $$n(S)$$ is:

We need $$|z - i| = |z + i| = |z - 1|$$.

Let $$z = x + iy$$.

From $$|z - i| = |z + i|$$:

$$\sqrt{x^2 + (y-1)^2} = \sqrt{x^2 + (y+1)^2}$$

$$x^2 + y^2 - 2y + 1 = x^2 + y^2 + 2y + 1$$

$$-4y = 0$$, so $$y = 0$$.

From $$|z + i| = |z - 1|$$:

$$\sqrt{x^2 + (y+1)^2} = \sqrt{(x-1)^2 + y^2}$$

With $$y = 0$$:

$$\sqrt{x^2 + 1} = \sqrt{(x-1)^2}$$

$$x^2 + 1 = x^2 - 2x + 1$$

$$2x = 0$$, so $$x = 0$$.

Therefore $$z = 0$$ is the only solution. Let's verify: $$|0 - i| = 1$$, $$|0 + i| = 1$$, $$|0 - 1| = 1$$. ✓

$$n(S) = 1$$.

The answer is $$\boxed{1}$$, which corresponds to Option (1).