Equivalently, we want to find a permutation $\{a_i\}$ of $[12]$ so that no two $(a_i+i)$'s are equal and no two $(a_i - i)$'s are equal.
This can be seen to be equivalent to placing $12$ queens on a $12 \times 12$ chessboard so that no two attack each other, which makes the problem easier to solve.
After some tinkering around, we stumble upon the permutation $(1, 6, 8, 10, 7, 4, 2, 9, 12, 3, 5, 11)$ which works. This means that the answer is $\boxed{yes}.$
$\square$
@below lol I think it's much easier to place $12$ non-attacking queens on a chessboard than directly find an $\{a_i\}$ that works