We have that $f^6(k)=k$,by plugging $f(f(f(f^3(k))))=f(f(f(g(k))))=g(g(k))=k$(it is easy to check that $g(g(k))=k$).Now,we will prove that for every $k$,the minimal $m$ for which $f^m(k)=k$ is $6$.That will mean that the graph of $f$ consists of disjoint $6$ cycles,from which we can conclude that $6|2n$,or $3|n$.Now,candidates for $m$ are $1,2$ and $3$.If $m=1$,then $f^3(k)=k$ which is never equal to $g(k)$.If $m=2$,then $f^3(k)=f(k)=g(k)$,which can't be true.If $m=3$,then $f^3(k)=k$ which is never equal to $g(k)$.This concludes the proof,we have that $f$ is composed from $6$-cycles and $3|n$.It is not hard to construct an example in case $n=999$ by making diametricaly opposite numbers in the cycle being $k$ and $g(k)$(remind that $g(g(k))=k$).Thus,we are done.