Is there a bijection between a square and circle that is (1) area-preserving (2) conformal (angle-preserving) on the perimeter (except at the corners)?
This map satisfies (1) but not (2). This theorem constructively proves the existence of map satisfying (2) but not (1).
Suppose $f$ is the desired map, and let $\mathbf J$ be its Jacobian matrix. On the boundary (except the corners), $f$ is conformal, so $\mathbf J$ is a multiple of a rotation matrix. But $f$ is also area-preserving, so $\det\mathbf J = 1$, so $\mathbf J$ actually is a rotation matrix. Therefore $f$ preserves lengths along the boundary. But this is impossible, because the boundaries of a square and a circle of equal area do not have the same length.