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Consider an odd chessboard with an odd number of squares. A king is placed on each square of the board, then the kings are picked up and placed on each square of the board again. Can this be done in a way that every king is in a square next to its original position?

Note: we do not follow the rules of chess in this question, it is practical

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No, because every pawn will change the color of the square it stays on, and there is different number of light and dark squares on the odd-sized board.

Edit: Adding a new point into the question is a bad practice, see https://meta.stackexchange.com/questions/43478/exit-strategies-for-chameleon-questions

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    If the board is even, it is possible to split it into pairs of adjacent squares and just swap the pawns on them, so the answer is positive in this case. – 2017-02-01
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    Prove what?​ – 2017-02-01