One result I like is that the number of 321-avoiding permutations of length 2n whose matrices are 180°-symmetric is (2n choose n). The best proof I know is fairly short, but I wouldn't call it bijective:
Under the Robinson-Schensted correspondence, the 180°-symmetric permutations are exactly the ones which map to ordered pairs of self-evacuating tableaux, which are in turn in bijection with ordered pairs of domino tableaux in the same shape. (See Stembridge) Now, if you look at the 2-row (since our permutations must be 321-avoiding) domino tableaux of size 2n, there are n+1 Ferrers shapes they can take, and they can be formed from those of size 2n-2 in a way satisfying the relation in Pascal's triangle, so the sum over all 2-row Ferrers shapes of the square of the number of domino tableaux of that shape is the sum of the squares of the binomial coefficients (2n choose i), yielding (2n choose n).
I've tried to "unpack" each of these steps into a simple bijection, but nothing's budged. Still, it seems like the kind of problem that someone else might be able to solve.
