Well I'm not faring too well with this problem, but perhaps this will be of use: Assuming such function exists (which could be the start of a proof by contradiction, or the start of a way to find such an f), f(f(f(n))) can be thought of as either
Code:
f( f(f(n)) ) = f(-n) or f(f( f(n) )) = -f(n)
So f(-n) = -f(n), meaning f(n) is an odd function.
Edits: Ah ok so I think the answer is yes, and there is a simple function we can define.
Here is how I got there.
Suppose f(n) = m.
Then from above we know f(-n) = -f(n) = -m.
We also know f(m) = -n.
And consequently f(-m) = n.
So we partition the integers into groups of four, basically, (not counting 0) and the simplest is {0}, {-2,-1,1,2}, {-4,-3,3,4}, ...
So we can write
$\displaystyle f(n) = \begin{cases}0\quad\quad\quad\ \ n = 0 \\ n+1\quad\ \ n\ \text{odd and positive} \\ n-1\quad\ \ n\ \text{odd and negative} \\ -n+1\quad n\ \text{even and positive}\\-n-1 \quad n\ \text{even and negative}\end{cases}$
which can probably be written in a much more concise manner, anyway that should do the trick.