Characterization of one-to-one conformal mapping from unit disc onto a square

I'm trying to solve the following exercise from Rudin's Real & Complex Analysis, chapter 14 exercise 22.

Suppose $f$ is a one-to-one conformal mapping of $U$ onto a square with center at $0$, and $f(0) = 0$. Prove that $f(iz) = if(z)$. If $f(z) = \sum c_n z^n$, prove that $c_n = 0$ unless $n - 1$ is a multiple of $4$. Generalize this: Replace the square by other simply connected regions with rotational symmetry.

The part about $c_n = 0$ is easy once the first part is proved. This is were I'm stuck. I suspect the generalization will become easier once I know how to prove the first part. Any help is appreciated.

Notes:

• This is not homework. I'm reading the text on my own.
• $U$ is the unit disc.

Hint: $g(z)=f^{-1}(if(z))$ is a conformal 1-1 map from $U$ onto itself fixing the origin. Try playing a little with this or something similar.
• This yields that $f^{-1}(if(z))=\lambda z$ for some constant $|\lambda|=1$. How do we show $\lambda = i$? – user122916 Dec 5 '14 at 10:03
• @mrf I'm sorry, I still don't understand why the second equality hold, it is equivalent to $if(z)=f(iz)$, since f is 1-1? – Kira Yamato Dec 11 '14 at 13:01