1. Let the complex number be $z=x+iy$ where $x,y\in\mathbb{R}$.

2. Interpret the given conditions geometrically: $|z-1|=|z+1|=|z-i|$ This means the point $P(x,y)$ is at equal distance from the points $A(1,0),\quad B(-1,0),\quad C(0,1).$

3. First use $|z-1|=|z+1|$ Squaring both sides: $|x+iy-1|^2=|x+iy+1|^2$ $(x-1)^2+y^2=(x+1)^2+y^2$ $x^2-2x+1=x^2+2x+1$ $-2x=2x$ $x=0$

   So $z$ lies on the imaginary axis.

4. Now use $|z-1|=|z-i|$ With $x=0$: $|-1+iy|=|i(y-1)|$ Squaring: $1+y^2=(y-1)^2$ $1+y^2=y^2-2y+1$ $-2y=0$ $y=0$

5. Therefore, $z=0$

6. Verify: $|0-1|=1,\quad |0+1|=1,\quad |0-i|=1$ All are equal, so this is indeed a solution.

7. Since the equations gave a unique point, the number of complex numbers is $1$

Therefore, the correct option is A.