As I said in your previous question, a LFT is uniquely determined by 3 points and their images. If $z_1, z_2, z_3$ are your inputs (pick 3 points on your first circle) and $w_1, w_2, w_3$ are your images of these (pick 3 points on your second circle). Be sure they are in the same order as you move around the circle, as I mentioned before that LFTs preserve order. Then, use the formula
$$\frac{(z_1 - z_3)(z_2 - z)}{(z_2 - z_3)(z_1 - z)} = \frac{(w_1 - w_3)(w_2 - w)}{(w_2 - w_3)(w_1 - w)}$$
and then solve for $w$ to get $w$ as a function of $z$. This formula is one you need to know and is useful in many, many situations, not just this one.
Consider first the case of a biholomorphic $f \colon \mathbb{D} \to \mathbb{D}$, where $\mathbb{D}$ is the unit disk.
If we define $a = f(0)$, then the Möbius transformation
$$T_a \colon z \mapsto \frac{z - a}{1 - \overline{a}\,z}$$
is an automorphism of the unit disk, and hence $g = T_a \circ f$ is an automorphism of the unit disk that fixes $0$.
Lemma (H. A. Schwarz): Let $h \colon \mathbb{D} \to \mathbb{D}$ a holomorphic function with $h(0) = 0$. Then $\lvert h(z)\rvert \leqslant \lvert z\rvert$ for all $z \in \mathbb{D}$, and $\lvert h'(0)\rvert \leqslant 1$. If there is a $z \in \mathbb{D}\setminus \{0\}$ with $\lvert h(z)\rvert = \lvert z\rvert$ or $\lvert h'(0)\rvert = 1$, then $h(z) = cz$ for some $c$ with $\lvert c\rvert = 1$.
Since $g$ and $g^{-1}$ satisfy the assumptions of the lemma, we conclude first
$$\lvert z\rvert = \lvert g^{-1}(g(z))\rvert \leqslant \lvert g(z)\rvert \leqslant \lvert z\rvert$$
for all $z \in \mathbb{D}$. Thus the last sentence of the lemma applies and hence $g(z) = cz$, so $g$ is a Möbius transformation. But then
$$f = T_a^{-1}\circ g$$
is also a Möbius transformation since the family of Möbius transformations is closed under composition and inverses.
Now, if $U,V$ are arbitrary disks or half-planes, and $f \colon U \to V$ is biholomorphic, consider the map $g = T \circ f \circ S$, where $S$ is a Möbius transformation mapping $\mathbb{D}$ biholomorphically to $U$, and $T$ is a Möbius transformation mapping $V$ biholomorphically to $\mathbb{D}$. Then $g$ is an automorphism of the unit disk. By the first part, $g$ is a Möbius transformation, and consequently $f = T^{-1}\circ g \circ S^{-1}$ is also a Möbius transformation.
Best Answer
If $T(1)=0$, take $T(z)=\dfrac{a(z-1)}{cz+d}$.
If, furthermore, $T(\infty)=1+i$, take $a=1+i, c=1$ and get $T(z)=\dfrac{(1+i)(z-1)}{z+d}$.
Now find $d$ such that $T(4)=1-i$.