How do you prove that the mean of the co-primes of a number is half the number? Say $n = 6$, The set of co-primes is $\{1, 5\}$, $\text{mean} = 3$
For $n = 9$, the set of co-primes is $\{1, 2, 4, 5, 7, 8 \}, \text{mean} = 4.5$

Question: Prove that the mean of co-primes of $n$, which are less than $n$ is half the number itself.

I computed all values until $10000$ and it seems to hold good.
 A: I'm posting a separate answer because I think Inceptio's is far too complicated for a problem this simple.
Note that $a$ is relatively prime to $n$ if and only if $n - a$ is relatively prime to $n$.  Also, generally $a \ne n - a$ (though there is one exception, which you have to deal with separately.  Can you find the exception?)
Anyway, this means we can make a list of everything relatively prime to $n$ in pairs of the form $(a, n - a)$, i.e. our list is:
$$
a_1, n - a_1, a_2, n - a_2, a_3, n - a_3, \cdots , a_k, n - a_k
$$
We can also guarantee that this list is all-inclusive (why?) and that no number appears twice (how?).
Then simply add up the numbers in the list and divide by the total.  You should get that the average is $\frac{n}{2}$.
A: Let $\{a_1, \dots a_{ \phi(n)} \}$ be set of all co-primes of $n$, you notice that $a_{\phi(n)}=n-1( Why?)$ and $a_1=1$. 
Let $n= p_1^{\alpha_1} \cdots p_k^{\alpha_j}$ ,  $p_i \nmid a_k$ where $1 \le i \le k$ and $1 \le j \le \phi(n)$
As rightly noted by Battacharjee in his comment, $\gcd({n,a_i})=1=\gcd(a_i,n-a_i)$
And the proof for that:
$(n,a_i)=d \implies n=df, a=dt \implies \gcd(dt,d(f-t))=d=\gcd(a_i,n)$,since $\gcd(t,f)=1$.
Here $\gcd(a_i, n)=1 \implies \gcd(a_i,n-a_i)=1$
Note that: $a_i, n-a_i \in \{a_1, \dots a_{ \phi(n)} \}$
$$\sum_1^{\phi(n)}a_i= \dfrac{\phi(n) n}{2}$$
Now the result is quite direct. 
$$\text{Mean}= \dfrac{\sum_1^{\phi(n)}a_i}{\phi(n)}=\dfrac{n}{2}$$
Some things that need to be explained:
$\phi(n)$ denotes the Totient function. Number of co-primes of $n$ , less than $n$. You should read more about it. 
$a_i, n-a_i \in \{a_1, \dots a_{ \phi(n)} \}$(Why?)
Because: $\gcd(a_i, n-a_i)=1$, and all numbers which are co prime to $n$ are in the set, which means they are in the set.
The answer looks too complicated only because of notations. Look at Goos' answer.
A: I assume $n\neq2$. The case $n=2$ should be treated separately (see Goos' comment).
Since $\gcd(a,n)=1\iff\gcd(n-a,n)=1$ it follows that if $C=\{a_1,a_2,\ldots,a_r\}$ is the set of the co-prime positive integers with $n$ (and smaller of $n$) then we can write $$C=\{a_1,a_2,\ldots,a_s,n\color{brown}{-a_s},\ldots,n\color{brown}{-a_2},n\color{brown}{-a_1}\}$$
for $s=\dfrac{r}{2}$.
 
In your examples for $n=6$, $C=\{1,5\}=\{1,6\color{brown}{-1}\}$ and
 
for $n=9$, $C=\{1,2,4,5,7,8\}=\{1,2,4,9\color{brown}{-4},9\color{brown}{-2},9\color{brown}{-1}\}$.
 
Therefore $\sum C=s\cdot n$ and the mean is $\dfrac{\sum C}{2s}=\dfrac{s\cdot n}{2s}=\dfrac{n}{2}$.
