user21899: From move 0, what shortest-possible sequence of moves could be played so that the pieces are arranged exactly as standard board setup (with allowed swapping of same-color knights), but it is black's turn to move?  Is it even possible?
I thought about it and conjectured white would have to make an odd number of knight moves, and black would have to make an even number of knight moves. I think this is impossible when the knights need to end up where they started, or even swap with each other (because even though this is an odd number of moves for one knight, the other knight would also have to make an odd number of moves). Am I correct? How can it be proven?
I also guess that this is a famous or at least previously explored problem. Maybe it's trivial.

user15509: You are basically trying to make a knight lose a tempo, which is known to be impossible.

user27836: This is impossible because.

In the initial position, the 4 knights cover 2 white and 2 black squares.
each white move will lead to the knights being on 3 white squares and 1 black square or 1 white and 3 black.
each other black move would lead to 2 & 2 or 4 & 0.

In order for "black to go first" the knights will need to be on the initial position (2 white, 2 black).
This is not possible after a white's move as shown above.

user3270: It's impossible. Knights require an even number of moves to reach a square of the same color as where they started, or correspondingly an odd number of moves to reach the opposite color. Both sides need to make an even number of moves to reach the starting position.

21899: <p>From move 0, what shortest-possible sequence of moves could be played so that the pieces are arranged exactly as standard board setup (with allowed swapping of same-color knights), but it is black's turn to move?  Is it even possible?</p>
<p>I thought about it and conjectured white would have to make an odd number of knight moves, and black would have to make an even number of knight moves. I think this is impossible when the knights need to end up where they started, or even swap with each other (because even though this is an odd number of moves for one knight, the other knight would also have to make an odd number of moves). Am I correct? How can it be proven?</p>
<p>I also guess that this is a famous or at least previously explored problem. Maybe it's trivial.</p>
 15509: <p>You are basically trying to make a knight lose a tempo, which is known to be impossible.</p>
 27836: <p>This is impossible because.</p>
<ul>
<li>In the initial position, the 4 knights cover 2 white and 2 black squares.</li>
<li>each white move will lead to the knights being on 3 white squares and 1 black square or 1 white and 3 black.</li>
<li>each other black move would lead to 2 &amp; 2 or 4 &amp; 0.</li>
</ul>
<p>In order for &quot;black to go first&quot; the knights will need to be on the initial position (2 white, 2 black).<br />
This is not possible after a white's move as shown above.</p>
 3270: <p>It's impossible. Knights require an even number of moves to reach a square of the same color as where they started, or correspondingly an odd number of moves to reach the opposite color. Both sides need to make an even number of moves to reach the starting position.</p>
