Equality holds in Hadamard's inequality iff $u(e_1),\ldots,u(e_n)$ are pairwise orthogonal.
Equality in the geometric-arithmetic mean inequality holds if and only if all eigenvalues of $w^*w$ are equal, i.e. excluding the trivial case $\l=0$ there is some $\l > 0$ such that $uv/\l=w/\l$ is an orthogonal mapping $u^\prime$, it follows that
$$
u(e_j)=uv(e_j)/d_j=u^\prime(e_j)/\l d_j~.
$$
Conversely if $u(e_j)$ are pairwise orthogonal, then the mapping $u^\prime:e_j\mapsto u(e_j)/\Vert u(e_j)\Vert$ is an isometry and since $|\det u^\prime|=1$:
$$
|\det u|
=|\det(u^\prime)|\prod_j\Vert u(e_j)\Vert
=\prod_j\Vert u(e_j)\Vert~.
$$