1. We are given

$$\omega = \frac{z}{z-\frac{i}{3}}$$

and

$$|\omega|=1.$$

We need the locus of $z$.

2. Using the modulus condition:

$$\left|\frac{z}{z-\frac{i}{3}}\right|=1$$

which implies

$$|z|=\left|z-\frac{i}{3}\right|.$$

3. Let

$$z=x+iy.$$

Then

$$|z|=\sqrt{x^2+y^2}$$

and

$$z-\frac{i}{3}=x+i\left(y-\frac13\right),$$

so

$$\left|z-\frac{i}{3}\right|=\sqrt{x^2+\left(y-\frac13\right)^2}.$$

Thus,

$$\sqrt{x^2+y^2}=\sqrt{x^2+\left(y-\frac13\right)^2}.$$

Squaring both sides,

$$x^2+y^2=x^2+\left(y-\frac13\right)^2.$$

So,

$$y^2=y^2-\frac{2y}{3}+\frac19.$$

Hence,

$$\frac{2y}{3}=\frac19$$ $$y=\frac16.$$

4. Therefore the locus is

$$y=\frac16,$$

which is a straight line.

5. Geometric interpretation: $|z|=\left|z-\frac{i}{3}\right|$ means the point $z$ is equidistant from $0$ and $\frac{i}{3}$, so it lies on the perpendicular bisector of the segment joining these two points — a straight line.

Therefore, the correct option is:

$$\boxed{\text{C: a straight line}}$$

6. Comparison with stored answer: Stored correct answer is C, which matches our result.