Shuffle cards into the original order except for two cards

Question

Suppose I have 16 cards ordered from 1 to 16, and I shuffle them using the standard interlacing shufflee (so if the original order was 1,2,...,16, the new order is 1,9,2,10,3,11,4,12,5,13,6,14,7,15,8,16 after one shuffle). Is it possible to shuffle some number of times so that the cards are in the original order except two consecutive cards have been switched (for example, 1, 2, 4, 3, 5, 6, ...)?

I know a permutation is needed for this but I don't know where to start.

Any help would be appreciated

Answer 1

This method of shuffling produces a cycle of length four:

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16
1,9,2,10,3,11,4,12,5,13,6,14,7,15,8,16
1,5,9,13,2,6,10,14,3,7,11,15,4,8,12,16
1,3,5,7,9,11,13,15,2,4,6,8,10,12,14,16
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16

These are the only four permutations achievable.

Answer 2

Consider the cycle structure of this permutation (which you can read off from the first two rows in Daniel Mathias’ answer):

$$ (1)(9\,5\,3\,2)(4\,10\,13\,7)(6\,11)(8\,12\,14\,15)(16) $$

First, since the cycle lengths all divide $4$, we can recover the order $4$ of the permutation as established by Daniel Mathias.

Second, there is an even number of cycles of even length. Thus, this is an even permutation. This alone tells us that none of its powers can be an odd permutation, and the transposition that you’re trying to reach is an odd permutation.