No-Luck Riomino

Question

Kris Burm's Riomino (or Tashkent Domino) is played with 25 identical dice that look like this:

The actual rules are as follows.

1. Each player rolls 12 dice that then serve as their "dominos" (once a die is rolled the face showing up must stay up for the remainder of the game).
2. The last die is rolled and placed between the players, who then take turns placing their "dominos" such that adjacent dots match (no other restrictions). The polyomino thus formed cannot take up more than five squares in either direction.
3. A player who cannot move loses.

Now when I took this game to Friday Night Magic yesterday with just the dice and no instruction manual, I forgot rule 1 above and inadvertently made the game of No-Luck Riomino:

1. The dice are common property. Starting from an empty playing field, each player takes turns to place a new cube in any orientation they like, but still obeying the dominoes and bounding box restrictions in rule 2 above.
2. A player who cannot move loses.

No-Luck Riomino is an impartial normal-play game, so one player has a winning strategy – but who?

Answer 1

No-Luck Riomino is a win for the

first player, by mimicking the second player.

Specifically,

the first player places his first die so that 0 and 2 (or 1 and 3) are showing. Then, after each move by the second player, the first player plays such that the play field remains unchanged after a 180-degree rotation around the first die and increasing each number by 2 (modulo 4).

Such a move will always be possible, because

the first player will never play adjacent to the second player's last move due to odd symmetry, the total number of dice is odd, and any change in bounding box initiated by the second player will always go from odd to even, and thus the mirror move by the first player will change it from even to odd (and therefore never from 5 to 6).