${}\qquad\qquad\qquad\dfrac{z-z_0}{-z_0 z+1}$WRONG ANSWER; SEE BELOW.
Or else
${}\qquad\qquad\qquad\dfrac{e^{i\theta}(z-z_0)}{-z_0 z+1}$WRONG ANSWER; SEE BELOW.
where $\theta$ is real. The first solution above is the case where $\theta=0$. If I'm not mistaken, the first solution above leaves the circle pointwise fixed. The other ones rotate it.
ERRATUM: I realized I mangled this five minutes after posting it, then couldn't get back here till now. I gave the unique l.f.t. that maps $z_0$ to $0$ and fixes $\pm1$. But I should have given the one that fixes $\pm z_0/|z_0|$. That's the one that would leave the circle invariant. But not pointwise fixed---that's silly. If you fix three points with this kind of mapping, then you fix all points.
continued....
So if we want to fix $\pm z_0/|z_0|$ we can do this: $ \pm\frac{z_0}{|z_0|} \mapsto \pm1 \mapsto \pm1 \mapsto \pm\frac{z_0}{|z_0|} $ where the first arrow is $z \mapsto z/\left(z_0/|z_0|\right)$ (so this takes $\pm z_0/|z_0|$ to $\pm1$ and takes $z_0$ to $|z_0|$), and the second arrow is the unique linear fractional transformation that fixes $\pm1$ and takes $|z_0|$ to $0$, and the third arrow takes $\pm1$ back to $z_0/|z_0|$. So the second arrow is $ \frac{z-|z_0|}{-|z_0|z+1}. $ The third arrow can be dispensed with, since it's just a rotation that leaves the circle invariant and leaves $0$ fixed. So we have $ \frac{\frac{z}{z_0/|z_0|} - |z_0|}{-|z_0|\left(\frac{z}{z_0/|z_0|}\right) +1}. $
How do we know this leaves the circle invariant? Because the mapping above, with the three arrows, leaves $\pm z_0/|z_0|$ fixed and maps $z_0$, which is on the line between those two points, to $0$, which is also on the line between those three points. So it's just like a mapping that fixes $\pm1$ and takes a real number to another real number. Mappings that do that are of the form $(az+b)/(bz+a)$ where $a$ and $b$ are real. You can check that this leaves the circle invariant by looking at $u+iv$ where $u,v$ are real and $u^2+v^2=1$ and putting it through this mapping and summing the squares of the real and imaginary parts and getting $1$. There's probably also a slick way to do it.