Show that Möbius transformation $S$ commute with $T$ if $S$ and $T$ have the same fixed point. Let $T$ be a Möbius Transformation such that $T$ is not the identity. Show that Möbius transformation $S$ commute with $T$ if $S$ and $T$ have the same fixed point. 
Here is what I know so far
1) if $T$ has fixed points says $z_1$ and $z_2$ then $S^{-1}TS$ has fixed point $S^{-1}(z_1)$ and $S^{-1}(z_2)$
2) if $T$ is dilation then $0$ and $\infty$ are its only fixed point, but if $T$ is translation then only $\infty$ is its fixed point.
Assume that $S$ and $T$ has the same fixed points $z_1$ and $z_2 $ then by 1)
$S^{-1}TS$ and $T^{-1}ST$ have the same fixed point $S^{-1}(z_1)=T^{-1}(z_1)$ and $S^{-1}(z_2)=T^{-1}(z_2)$
I know that $T$ is not the identity, but I can't assume it is dilation or translation to use 2), because it can also be inverse, right?
I wonder if anyone would please have me a hand from here.

I thought about this the other day. 
Suppose that 
$M_1(z) = \frac{a_1z + b_1}{c_1z + d_1}$ and $M_2(z) = \frac{a_2z + b_2}{c_2z + d_2}$ 
are Möbius transforms with the same two fixed points $z_0$ and $z_1$. 
Then $\;c_1 z^2 + (d_1 - a_1)z - b_1 = c_1(z-z_0)(z-z_1)\;$ and
$\;c_2 z^2 + (d_2 - a_2)z - b_2 = c_2(z-z_0)(z-z_1).$
With $c_1 = kc_2$ for some $k,$ we get $\;b_1 = -kc_2 z_0 z_1 = k b_2 ~~~~~(*),\;$
and $\;d_1 - a_1 = -kc_2 (z_0+z_1) = k(d_2 - a_2) ~~~~~ (**)$.
Now, $M_1(M_2(z)) = \frac{(a_1a_2 + b_1c_2)z + a_1b_2 + b_1d_2}{(c_1a_2 + d_1c_2)z + c_1b_2 + d_1d_2}$ 
[Using (*)]
$=  \frac{(a_1a_2 + kb_2c_2)z + b_2(a_1+kd_2)}{c_2(ka_2+d_1)z + c_2b_1 + d_2d_1}$ 
[Using (**) and (*)]
$= \frac{(a_2a_1 + b_2c_1)z + b_1a_2+ b_2d_1}{(c_2 a_1 + c_1d_2)z + c_2b_1 + d_2d_1} = M_2(M_1(z)).$  
