Let $f(z)=(az+b)/(cz+d)$ and $g(z)=az/(ez+f)$ be two Möbius transformations, with $a,b,c,d,e,f$ real numbers (note the same coefficient $a$ in $f$ and $g$) and with $cz+d$ and $ez+f$ non constant.
My question is: What conditions we can deduce about the coefficients of $f$ and $g$, if $f(\omega)=g(\omega)$, for some non-real number $\omega$?
On
Well, you can just write out what $f(\omega)=g(\omega)$ means: $$\frac{a\omega+b}{c\omega+d}=\frac{a\omega}{e\omega+f}.$$ Cross-multiplying and simplifying gives $$a(e-c)\omega^2+(be+a(f-d))\omega+bf=0.$$ If the leading coefficient $a(e-c)$ is nonzero, this is a quadratic with real coefficients, so it has a nonreal root iff the discriminant $$(be+a(f-d))^2-4abf(e-c)$$ is negative.
If $a(e-c)=0$ then the only way there can be a nonreal root is if all the coefficients are $0$. This just means that $f(z)=g(z)$ for all $z$, so they have the same coefficients up to scaling and hence the same coefficients since they have a common nonzero coefficient of $a$. (Note that if you actually try to solve for when $a(e-c)=be+a(f-d)=bf=0$, you will find some other solutions such as $a=b=0$, but those are all degenerate and would make the original functions not actually be Möbius transformations.)
You have two functions which take the same value at one point, where this point has non-trivial imaginary part (if I understand you correctly). This might impose some restriction on $a,b,c,d,e,f$, for example you can express $f$ as a function of your other five constants, but that's probably it.