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I've been trying to solve this question
It seems that in order to show it has unit length, we must show that $$ \frac{1}{d} \sum_{m, n=0}^{d=1} \lvert U_{m, n}\rvert ^2 = 1 $$
I've tried searching online for this type of relation on unitary matrices but haven't succeeded in finding anything. Would appreciate any help.
It may be more helpful to think about the relationship you need to show in terms of the Hermitian form $\sum \limits_{m,n=0}^{d-1} U_{m,n} U_{m,n}^\ast = d$, which is a necessary condition for the defining unitary relationship $UU^\dagger = I$.
To see this explicitly, consider the four equations implicit in the $d=2$ case: $$UU^\dagger=\begin{bmatrix} U_{1,1} & U_{1,2} \\ U_{2,1} & U_{2,2} \end{bmatrix} \begin{bmatrix} U_{1,1}^\ast & U_{2,1}^\ast \\ U_{1,2}^\ast & U_{2,2}^\ast \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$ The diagonal terms are what you need for the desired relationship: $$(U_{1,1} U_{1,1}^\ast + U_{1,2} U_{1,2}^\ast) + (U_{2,1} U_{2,1}^\ast + U_{2,2} U_{2,2}^\ast) = 1 + 1 = 2.$$ It should not be hard to convince yourself that this generalizes to any $d$.
A matrix is unitary iff its columns (equivalently, rows) are orthonormal (see e.g. Wikipedia).
The sum $\sum_{nm}|U_{nm}|^2$ can be thought of as the sum of the squared norms of the rows of $U$ (equivalently, of the columns of $U$). As each such row (column) has unit norm, the sum of these norms must equal $d$.
Being $U$ unitary, its SVD has the form $U=\sum_k |u_k\rangle\!\langle k|$ for some orthonormal basis $(|u_k\rangle)_k$. The corresponding state $|\Psi\rangle$ thus reads $$|\Psi\rangle = \frac{1}{\sqrt d}\sum_{k=1}^d |u_k\rangle\otimes|k\rangle,$$ which is the maximally entangled state.
