Given some ordered list of $n$ items, I obviously have $n!$ possible arrangements. However, as a heuristic in some larger algorithm that needs to find the optimal order, I want to do some quick local exploration. One option is to walk through all possible removal & insert operations at $O(n(n-1))$. A similar option is to run through all possible 2-swaps (e.g 2-opts), also at a cost of $O(n(n-1))$. The question: do those two heuristic approaches cover the same solution set? Are they equivalent?