Find a Mobius transformation $\phi$ that maps $\{z\in\mathbb{C}\,|\,\text{Im}(z)>0\}$ to $\{z\in\mathbb{C}\,|\,\text{Im}(z)<0\}$, such that $\phi(i)=-i$ and $\arg(\phi'(i))=-\dfrac{\pi}{2}$.
I tried to say that $\phi(z)=\dfrac{az+b}{cz+d}$ and then $ai+b=c-di$.
Then, the other condition gives us that $\text{Re}(\phi'(i))=0$ and $\text{Im}(\phi'(i))<0.$
So, $\phi'(z)=\dfrac{ad-bc}{(cz+d)^2}$, so $\phi'(i)=\dfrac{ad-bc}{(ci+d)^2}$.
The moebius transformation we are looking for maps the real axis to itself, hence the coefficients $a$, $b$, $c$, $d$ may be taken to be real. The condition $ai+b=c-di$ then enforces $b=c$ and $a=-d$. This leaves us with $$T(z)={az+b\over bz-a}\ .$$ Since the coefficients are only determined up to a real nonzero factor we may write $$T(z)={z\cos\alpha +\sin\alpha\over z\sin\alpha -\cos\alpha}\ .\tag{1}$$ This gives $$T'(i)={-\cos^2\alpha-\sin^2\alpha\over\bigl(-(\cos\alpha-i\sin\alpha)\bigr)^2}={-1\over e^{-2i\alpha}}=-e^{2i\alpha}\ .$$ Since we want ${\rm arg}\bigl(T'(i)\bigr)=-{\pi\over2}$ we have to choose $\alpha={\pi\over4}$ (or $={\pi\over4}+\pi$, which leads to the same $T$). Plugging this into $(1)$ and cancelling the $\sqrt{2}$ everywhere then gives $$T(z)={z+1\over z-1}\ .$$