So what we have to work with is that we know $f(z_0) = z_0$, and that our conjugacy function is $\phi (z) = z+z_0$. We want to show that the conjugate $g(0)=0$.

The way I started this problem was setting up our conjugation, namely $f(\phi(z))=\phi(g(z))$ and substituting in what I know.

$\phi(g(z)) = g(z) + z_0$,

$f(\phi(z)) = f(z+z_0)$

We want to show that when $f(z_0) = z_0, g(0)=0$.

Substituting $z=0$ into the two equations, we get $g(0) + z_0 = f(0+z_0)$.

This turns into $g(0)+z_0 = f(z_0)$, and we know $f(z_0) = z_0.$

Thus, $g(0) +z_0 = z_0$, and after subtracting a $z_0$ 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 $f'(z_0)$ is related to $g'(z_0)$, but here's where I'm stuck.

HELP