Or stated simply, what is the arithmetic mean of the totatives of $n$?
From this question here I can see that the sum of the totatives is given by the formula $\large\frac{n\times\phi (n)}{2}\large$. So the required answer would then be $\frac{n}{2}$. What is a simpler or an alternative argument to prove this?
Easier way: If $k$ is relatively prime to $n$, so is $n-k$. So the numbers in $S$ can be paired up so that the sum of each pair is $n$.