Find the Möbius transformation that...

With $f(z)=\frac{az+b}{cz+d}$ we have that $$ f(\infty)=\frac{a}{c}=1, \qquad f(-1)=\frac{-a+b}{-c+d}=i, \qquad f(i)=\frac{ai+b}{ci+d}=1+i. $$ Thus, we need to solve the three equations with respect to four unknowns, where one of them can be chosen freely.

We can find $\omega(z)=\dfrac{az+b}{cz+d}$ using formula $$\frac{z-z_1}{z-z_2}:\frac{z_3-z_1}{z_3-z_2}=\frac{\omega-\omega_1}{\omega-\omega_2}:\frac{\omega_3-\omega_1}{\omega_3-\omega_2}$$ where $(z_1,z_2,z_3)=(-1,\infty,i)$ and $(\omega_1,\omega_2,\omega_3)=(i,1,1+i).$

In our case we have $$\frac{z+1}{1}:\frac{i+1}{1}=\frac{\omega-i}{\omega-1}:\frac{1}{i},$$ where we change $z-z_2$ with $1$ because $z_2=\infty,$ etc.

And now we find that $\omega=\omega(z)=\dfrac{z+2+i}{z+2-i},$ like previous solutions.

Suppose the Möbius transformation is that $M(z)=\frac{z+a}{z+b}$, and then by $M(-1)=i,M(i)=1+i$, we can get that $a=2+i,b=2-i$, so the desired Möbius transformation is that $$M(z)=\frac{z+2+i}{z+2-i}.$$