Q: Find a mobius transformation T with the following property or show that no such T exists.
There are distinct points $z_1, z_2, z_3, a, b$ in $\mathbb{C} \cup \{\infty\}$ with $T(z_1)=z_2$, $T(z_2)=z_3$, $T(z_3)=z_1$ and $T(a)=b$, $T(b)=a$?
By the theorem, there exists a unique mobius map that maps three distinct points to three distinct points. So there exists a map $T$ such that $T(z_1)=z_2$, $T(z_2)=z_3$, $T(z_3)=z_1$ and actually this $T$ maps a circle(or extended line) to itself.
Let $T$ be a linear fractional transformation such that for distinct $z_1,z_2,z_3$ in the "extended complex plane" $\mathbb{C} \cup \{\infty\}$ (also known as the Riemann sphere):
$$ T(z_1) = z_2\;, T(z_2) = z_3\;, T(z_3) = z_1 $$
Since $T^3(z_i) = z_i$ for $i=1,2,3$, it follows that $T^3$ is the identity since agreement of two linear fractional transformations on three distinct points implies they are the same.
Assume further that $a,b$ are such that $T(a) = b$ and $T(b)=a$. Then:
$$ a = T^3(a) = T^2(b) = T(a) = b $$
so $a,b$ cannot be distinct.