Extending an automorphism to the integral closure I need some help to solve the second part of this problem. Also I will appreciate corrections about my solution to the first part. The problem is the following.

Let $\sigma$ be an automorphism of the integral domain $R$. Show that $\sigma$ extends in an unique way to the integral closure of $R$ in it's field of fractions.

My solution: denote $\overline{R}$ for the normalization of $R$ and $\overline{\sigma}$ for the extension of $\sigma$. Given some $r/s\in \overline{R}$, we can define $\overline{\sigma}(r/s) = \sigma(r)/\sigma(s)$. From this I can show that $\overline{\sigma}(r/s)\in\overline{R}$. So this looks like a decent extension. I guess the extension is supposed to be a homomorphism (or even an automorphism), althought this is not said explicity.
The problem comes when I try to show it is unique. I don't know what to do, all my approaches failed. 

The uniqueness is fairly trivial. Since suppose $\bar{\sigma}$ is an extension then it must be defined as you said. Indeed $\bar{\sigma}(r/s)=\bar{\sigma}(r\cdot s^{-1})=\bar{\sigma}(r)\cdot\bar{\sigma}(s)^{-1}=\sigma(r)\cdot\sigma(s)^{-1}=\sigma(r)/\sigma(s)$. And indeed $\bar{\sigma}$ is an automorphism if $\sigma$ is since the inverse is given by the unique extension of $\sigma^{–1}$.
