The game begins with $n$ sets of brackets in the outer layer.
[][][]
Players then take turns choosing a pair of brackets, then placing it and all its contents into a different bracket in the same layer.
[][][] -> [][[]] -> [[[]]]
Another example; From [()][()], the only move available is [()[()]] (since order does not matter). Note round brackets are equivalent and are used for easier reading.
The person who cannot move loses.
As shown in the first example, $n=3$ is a win for the second player. $n=4$ is also a win for the second player:
[][][][] -> [[]][][] -> [[]][[]] -> [()[()]] -> [[[[]]]]
$n=5$ is also a win for the second player, but beyond that it is difficult to check by hand.
I think that all $n\ne 2$ are a win for the second player.
If we take any position, and replace each bracket with a pair of brackets, then the position becomes a win for the second player. For example, all of [([()][()])], [()][()][([()])], etc are a win for the second player. I call these positions 'double positions'.
Proof: The first player must choose an 'outer bracket', which are shown as [], and place it inside another outer bracket. The second player can take the same bracket and put it in the inner bracket, which are shown as (). This results in another 'double position', meaning that the second player is guaranteed to always have a move.
Another observation I have made is related to stacks. Call a group of brackets an $n$-stack if there are exactly $n$ brackets with no moves available. For example:
1-stack - []
2-stack - [[]]
3-stack - [([])]
and so on. Take a position composed of two stacks of size $a$ and $b$. If $a=1$ or $b=1$, then the position is clearly a win for the first player. Otherwise, it is a win for the first player if and only if $a+b$ is odd.
This is because placing a stack inside another stack essentially reduces the size of that stack by $1$ (the outer bracket is irrelevant since it cannot be moved). This means both players will continue moving, avoiding a stack of size $1$ until there are two $2$-stacks, at which point the next player looses. But the parity of the sum of the sizes of the stacks changes after each step, so if $a+b$ was odd, then the first player will never reach [()][()].
Is anything known about this game? For which $n$ does the second player win? As stated above I think its $n\ne 2$ but I don't know how to prove it. Also, how does combining two positions affect the winner?