On p.32 of Carleson/Gamelin Complex dynamics they show existence of a conjugation map for a repelling fixed point using the attracting case and I am trying convince myself of the statement "any map conjugating $f^{-1}(z)$ to $\xi/\lambda$ also conjugates $f(z)$ to $\lambda \xi$. "
Let $f(z)=z_0+\lambda(z-z_0)+\dots$ and $f^{-1}(z)=z_0+(z-z_0)/\lambda+\dots$.
If $f^{-1}(\phi(z))=\phi(z)/\lambda$, why does it follow that $f(\phi(z))=\lambda\phi(z)$
I get $f(\phi(z))=f(\lambda f^{-1}(\phi(z))$ but $f$ is not linear so I cannot say this equals $\lambda \phi(z)$.
I don't think you've written your conjugation formula correctly. I think it should be $$ \varphi(f^{-1}(z)) = \frac{1}{\lambda}\varphi(z). $$
If we now write $w=f^{-1}(z)$, then $$ \varphi(w) = \frac{1}{\lambda} \varphi(f(w)) \: \text{ so } \: \varphi(f(w)) = \lambda \varphi(z). $$ That is $\varphi$ conjugates $f(z)$ to $\lambda z$.