We can identify the group of automorphisms of the upper half plane with $PSL(2,\mathbb{R})$, the group of $2\times 2$ matrices with determinant 1, up to $\pm I$. That is, they are transformations of the form
$$z \rightarrow \frac{az + b}{cz + d},$$
where $ad - bc = 1$ and $(a,b,c,d) \sim (\alpha a, \alpha b, \alpha c, \alpha d)$ for $\alpha \neq 0$, $a, b, c, d \in \mathbb{R}$. Observe that this is a three-parameter family.
If there were two such automorphisms $\Phi_1$ and $\Phi_2$, then $\Phi_1^{-1}\circ\Phi_2(x_j) = x_j$ for $j = 1, 2, 3$, i.e. there are three fixed points. The only automorphism which has three fixed points is the identity, so $\Phi_1^{-1}\circ\Phi_2 = I$. (Why? If $(az+b)/(cz+d) = z$ is a fixed point, then unless $c = a = 1$ and $b = d = 0$, we obtain at best a quadratic with at most two roots, by the fundamental theorem of algebra.)
$arg({a \over b})=arg(a)-arg(b)$
So your problem becomes...
$arg(z_3-z_2)-arg(z_3-z_1)=\frac{1}{2}(arg(z_2)-arg(z_1))$
Now let $z_k=re^{i\theta_k}=r(\cos{\theta_k+i\sin{\theta_k}})$
$arg(r(\cos{\theta_3}-\cos{\theta_2}+i(\sin{\theta_3}-\sin{\theta_2})))-arg(r(\cos{\theta_3}-\cos{\theta_1}+i(\sin{\theta_3}-\sin{\theta_1})))=\frac{1}{2}(arg(r(\cos{\theta_2+i\sin{\theta_2}}))-arg(r(\cos{\theta_1+i\sin{\theta_1}})))$
$arctan({\sin{\theta_3}-\sin{\theta_2} \over \cos{\theta_3}-\cos{\theta_2}})-arctan({\sin{\theta_3}-\sin{\theta_1} \over \cos{\theta_3}-\cos{\theta_1}})={1 \over 2}(arctan({\sin{\theta_2} \over \cos{\theta_2}})-arctan({{\sin{\theta_1} \over \cos{\theta_1}}}))$
It can be proven with trig identities that ${{\sin{x}-\sin{y}} \over {\cos{x}-\cos{y}} }={\cot({\frac{x+y}{-2}})}$
$arctan({\cot({\frac{\theta_3+\theta_2}{-2}})})-arctan({\cot({\frac{\theta_3+\theta_1}{-2}})})={1 \over 2}(arctan(\tan{\theta_2})-arctan(\tan{\theta_1})$
It can be shown that $arctan(\cot{x})={\pi \over 2}-x$
$({\pi \over 2} - \frac{\theta_3+\theta_2}{-2})-({\pi \over 2} - \frac{\theta_3+\theta_1}{-2})={1 \over 2}({\theta_2}-{\theta_1})$
$\frac{\theta_3+\theta_2}{2}-\frac{\theta_3+\theta_1}{2}={1 \over 2}({\theta_2}-{\theta_1})$
$\frac{\theta_3+\theta_2-\theta_3-\theta_1}{2}={1 \over 2}({\theta_2}-{\theta_1})$
$\frac{\theta_2-\theta_1}{2}={1 \over 2}({\theta_2}-{\theta_1})$
The statement appears to be proven, but be aware that this makes many assumptions, largest of which is that the angles all exist in the first quadrant. If the angles were all in different quadrants, the proof would be much more involved, and to make such a proof would need a lot of meticulous calculations and studying many different cases - mostly because of the arctan function and its limited range.
Good luck.
Best Answer
The idea is to show that
$T(z; x_1, x_2, x_3) = \dfrac{(z-x_1)(x_2-x_3)}{(z-x_3)(x_2-x_1)}$ is an automorphism of $\Bbb H$ which maps $x_1, x_2, x_3$ to $0, 1, \infty$, respectively, and
the automorphisms of $\Bbb H$ form a group.
Therefore $$ T(z; y_1, y_2, y_3)^{-1} \circ T(z; x_1, x_2, x_3) $$ is an automorphism of $\Bbb H$ with the desired property.
Remark: $\dfrac{(z-x_1)(x_2-x_3)}{(z-x_3)(x_2-x_1)}$ is, up to some permutation of the arguments, the so-called “cross-ratio” or “double ratio“ of $z, x_1, x_2, x_3$.