EDIT: My apologies. I got the picture.

If $n=4$ the first player will loose other wise he always has a winning strategy: For $n=3,5$ there is a unique way of playing up to a symmetry. For $n=7$, the winning strategy will be $7,6,3,$... if we label the edges from 1 to 7 cyclicaly. Now we prove by induction from $n$ to $n+3$ as follows: If we label the edges from $1$ to $n+3$ the strategy will be $n+3,n+2,n$ and then erase the edges from $n$ to $n+3$ and replace them by an edge $n$ and play the winning strategy for $n$.