Reiterate problem
Bunny: a chess piece which moves like a bishop but only one square from its current position. It may also hop over another piece.
Hop: a bunny hops when another piece is in a square touching the current square (no diagonals). In its turn it occupies the square 3 squares from its current square in the direction of the hopped piece. The hopped piece then takes the original square of the bunny. This is not considered a turn/move for the hopped piece.
Say, there are 2 bunnies on a board. $\alpha$ and $\beta$.
$\alpha$ and $\beta$ begin on opposite colour squares of your choosing.
By what means might they both tour the board? No turn may be taken to a previously occupied square (being hopped does not constitute occupation). They take turns, $\alpha$ moves first, $\beta$ second, and they will take 63 turns each.
Vocabulary
The bunny thus has 2 distinct modes of motion: the "step" and the "hop". If a piece is hopped over its motion (though not counted as a move) is a "pivot". A sequence of "step"s is a "walk".
When $\alpha$ moves to a square he paints the square $green$. When $\beta$ moves to a square he paints it $blue$. After both bunnies have moved to a square it is painted $red$ - and hereafter no bunny may enter except by "pivot", and then only immediately as the last bunny has moved to this square.
$Lemma\ 1:$ It is a known fact that no tour purely constistuted of steps is possible. In fact, it is known that it takes 4 distinct walks (sharing no squares) to cover all the black (or white) squares on the board.
$Lemma\ 2:$ As a corollary to the above, there are only a subset of total squares of a given colour reachable from any arbitrary square - where reachable means that it can be reached by only using some sequence of steps (no hopping). If all squares or a colour were reachable then $Lemma\ 1$ would be false.
Answer
I will now claim that this bunny's tour is impossible.
For $\alpha$, given any starting square $s$ on colour $c$, we know that $s$ must be a square on only one of the four walks needed to cover the square of colour $c$ - call this walk $w$. Any square visited from $s$ must be part of $w$. It is clear that his tour cannot be completed from $s$ ($Lemma\ 1$).
This means that $\alpha$ must escape and continue by means of a hop.
Now, $\beta$ occupies one of the squares in $w$, $s^{\prime}$. $\beta$ may now process on his walk $w^{\prime}$. We know that only the same squares that were reachable to $s$ are reachable to $s^{\prime}$. This means that they are part of the same walk. Eventually $\beta$ will get stuck or otherwise need to change to a different walk to have a chance at covering the board (as we know that 1 is not sufficient). His only means of escape is a hop. Now $\alpha$ takes a square on $w^{\prime}$, BUT $w^{\prime}$ has exactly the same reachable squares as $w$. So, $\alpha$ has gained no new reachable squares for colour $c$ ($Lemma\ 2$). And there is nothing he can do to remedy this situation.
Thus, there is no 2-bunny tour of a chessboard.
early observations for historical reasons only
I'm quite positive that there is no solution.
Let's say we have 2 bunnies, a black one and a white one. The black bunny paints the board with green paint, and the white bunny paints the board with blue paint. A square that has been painted with both green and blue paint, is red.
- a white bunny can no longer visit a blue square
- a black bunny can no longer visit a green square
- a red square can no longer be accessed by any bunny.
The answer to this question suggests that there is no "walking-bishop" tour. The bunny is a walking-bishop with a hopping capability.
Each bunny must visit every square. This means that at least 8 hops would be required. 3 squares are "involved" in a hop. And one of them will be guaranteed to be red after a hop. This means that the 8 hops will involve cancelling out at least $\frac{1}{8}^{th}$ of the entire board.
- they must change from black to white squares 4 time in order to get out of where they are stuck and to finish painting the board.
The dead-locks occur under the following conditions: (the first 3 are because the bunny's counterpart cannot each a hopping square)
- a white bunny is anywhere where the 4 opposite colour squares surrounding him are coloured green
- a black bunny is anywhere where the 4 opposite colour squares surrounding him are coloured blue
- any bunny is anywhere where the 4 opposite colour squares surrounding him are coloured red
- a bunny is on a red square and cannot move to non-red square without making it red (now hopping is not possible - as the other bunny would need to occupy a red square - i.e. a square he has already occupied in the past)
I will turn this into something solid soon.