The answer is no, and the basic idea is to look at parity. WLOG the compass has odd length; if not, simply scale everything down by the appropriate power of two. Then the parity of the sum of the coordinates of the two endpoints of the compass are different. Now mark the even-sum endpoint of the compass with a red dot, and the odd-sum endpoint with a blue dot. It's not terribly hard to see that no matter how many rigid rotations about one endpoint we make, the blue dot always occupies a lattice point with even coordinate sum, and the blue dot a point with odd coordinate sum. In particular, it's never possible to interchange the endpoints of the compass making moves of only this kind.