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There are two white and two black knights on a $3\times3$ chessboard:

enter image description here

Is it possible to get the following order of knights:

enter image description here

If knights on the first column are not moved, then we can't have the order in the second image. How can we check if this is possible by not checking all cases with all four knights?

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It is impossible. Think of a knight's 'tour' around this board: it takes 8 steps to go around clockwise or counterclockwise. Let's just focus on one of those tours, e.g. the one starting at the left top, and making the first step to the middle cell on the right.

If we now record what is in those cells that we encounter, we get:

B_W_W_B_

removing the blanks, we get BWWB. Call this the 'relative order' of the knights relative to this tour.

Now, as the knights move along this tour (think of it as race cars going around a track ... but they can go backwards as well) they can change their location in this tour, but the relative order will always remain BWWB (or, as they complete a 'lap' going forward or backward, WWBB or BBWW or WBBW), since they can never 'pass' each other along this tour, since that would mean that at some point they are in the same cell, which is of course not allowed. So, they can never get to the goal, whose knights will always have a relative order relative to that tour of BWBW or WBWB!

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    Rephrasing of this argument in graph language: the board can be represented as a closed graph: $1\to 6\to 7\to 2 \to 9 \to 4\to 3 \to 8 \to 1$ (number here is the number of the square counting squares from top to bottom and skipping the center $5$ as we can never get there). In this formulation the knight can only moves one step to the left or right on this graph for which the order you introduce is easily seen to be preserved. – 2017-01-22