Assuming rotations/reflections don't count, $f(n) = 1$ for all $n$ that can be written in the form $2^a + 2^b, a, b \in \mathbb{Z}^{+}$ and $0$ for all other $n$. From USAMO 2008 #4 (http://www.artofproblemsolving.com/Wiki/index.php/2008_USAMO_Problems/Problem_4), we have $f(n) > 0$ for these $n$ and $f(n) = 0$ for all other $n$. We also get from the proof of that problem that one diagonal must be drawn splitting the polygon into arcs of length $2^a$ and $2^b$, and from there on, all other diagonals are fixed. Discounting rotations/reflections, there is exactly one way to do so, so $f(n) = 1$ for these $n$. If rotations/reflections count, $f(n) = n$ when $a \neq b$ and $f(n) = \frac{n}{2}$ otherwise for these $n$.