I labeled a few lines and thought about rotating a three dimensional shape whose shadow is the picture below. I'm rotating counter-clockwise.
For two lines, labeled $1$ and $2$ I found that it takes $3$ moves to get back to the original configuration.
For three lines, I found that it takes $7$ moves to recover the initial setup.
For four lines, I found that it takes $13$ moves to recover the initial configuration.
Q: Is there a pattern for the sequence? I'd like to be able to calculate the nth term. So far I have $3,7,13.$
So for $n=3$ lines the full representation would consist of these $7$ steps:
$(1,2,3)\to(1,3,2)\to(3,1,2)\to(3,2,1)\to(3,1,2)\to(1,3,2)\to(1,2,3)$
For $n=2$ the $3$ steps would be:
$(1,2)\to(2,1)\to(1,2)$
If I understand your question correctly the solution you desire is $n(n-1)+1=n^2-n+1$ steps for $n$ lines. This is because every one of the $n$ lines passes every other one of the $(n-1)$ lines during a single rotation and the first step has to be added. For example we have: $$\begin{array} {|r|r|} \hline n&f(n)\\ \hline 1&1\\ \hline 2&3\\ \hline 3&7\\ \hline 4&13\\ \hline 5&21\\ \hline \end{array}$$