So what we have to work with is that we know f(z0)=z0, and that our conjugacy function is ϕ(z)=z+z0. We want to show that the conjugate g(0)=0.
The way I started this problem was setting up our conjugation, namely f(ϕ(z))=ϕ(g(z)) and substituting in what I know.
ϕ(g(z))=g(z)+z0,
f(ϕ(z))=f(z+z0)
We want to show that when f(z0)=z0,g(0)=0.
Substituting z=0 into the two equations, we get g(0)+z0=f(0+z0).
This turns into g(0)+z0=f(z0), and we know f(z0)=z0.
Thus, g(0)+z0=z0, and after subtracting a z0 from both sides we get g(0)=0.
To show that the nature of the fixed point is preserved, I think we need to show that f′(z0) is related to g′(z0), but here's where I'm stuck.
HELP