Let $H = (h_{ij})$ be a square matrix of order $n$ such that $|h_{ij}| \leq 1$. Then, by the Hadamard Determinant inequality we know that $$|\det(H)| \leq n^{\frac{n}{2}}$$
I read in this paper that, equality is achieved if and only if $H$ is a Hadamard matrix. The if is easy to show; as: $$HH^T = nI \iff \det(H) \det(H^T) = \det(nI) \iff \det(H)^2 = n^{n} \iff |\det(H)| = n^\frac{n}{2}$$
How do you go about showing the only if?
Assume that $|det(H)|=n^{n/2}$. Let $\sigma_i$ be the singular values of $H$.
We have $\det(H)^2=\det(HH^T)=\prod_i \sigma_i^2$, therefore $\prod_i \sigma_i^2=n^n$. By the AM-GM inequality, $$n=(\prod_i \sigma_i^2)^{1/n}\leq {\frac 1 n}\sum_i \sigma_i^2$$
On the other hand, $${\frac 1 n}\sum_i \sigma_i^2={\frac 1 n}Tr(H^TH)={\frac 1 n}\sum_{ij}h_{ij}^2\leq {\frac 1 n}\sum_{ij}1=n^2/n=n$$
Combining with the first inequality, we see that
$${\frac 1 n}\sum_i \sigma_i^2=\left(\prod_i\sigma_i^2\right)^{1/n}$$
Thus, the AM-GM inequality is actually an equality, which implies that $\sigma_1=\sigma_2=\dots=\sigma_n$.
This in turn implies that $H/\sqrt{n}$ is an orthogonal matrix. In particular, each column of $H$ must have norm $\sqrt{n}$, and under the constraint $h_{ij}^2\leq 1$, this can only be achieved when the entries are $\pm 1$.
Thus, $H$ is a matrix with $h_{ij} = \pm 1$ with mutually orthogonal columns, or a Hadamard matrix.
$\def\tr{\mathrm{tr}}$ For a $n \times n$ matrix $A$ we can write, $$ \epsilon^{i_1i_2\ldots i_n} {A^{j_1}}_{i_1} {A^{j_2}}_{i_2} \ldots {A^{j_n}}_{i_n} = \det(A) ϵ^{j_1j_2\ldots j_n} $$ (see Pavel Grinfeld, "Introduction to Tensor Analysis and the Calculus of Moving Surfaces", Chapter 9) where $\epsilon^{i_1i_2\ldots i_n}$ is the permutation symbol defined as, $$ \epsilon^{i_1i_2\ldots i_n}= \begin{cases} 1 & \text{$i_1i_2\ldots i_n$ is an even permutation of $123\ldots n$} \\ -1 & \text{$i_1i_2\ldots i_n$ is an odd permutation of $123\ldots n$} \\ 0 & \text{otherwise} \end{cases} $$ ${A^{j}}_i$ is the element of $A$ for row $i$ column $j$, ${A_j}^i$ denotes the matrix transpose and ${A^j}_i {A_j}^k$ denotes the matrix product $AA^T$. The only information we have for this matrix is that $\det(A)=n^{\frac{n}{2}}$. Lets write, $$ \det(A) ϵ^{j_1j_2\ldots j_n} \det(A) ϵ_{j_1j_2\ldots j_n} = \epsilon^{i_1i_2\ldots i_n} {A^{j_1}}_{i_1} {A^{j_2}}_{i_2} \ldots {A^{j_n}}_{i_n} \epsilon_{i_1'i_2'\ldots i_n'} {A_{j_1}}^{i_1'} {A_{j_2}}^{i_2'} \ldots {A_{j_n}}^{i_n'} $$ (both sides of the equation evaluate to a scalar; $\epsilon$ is a pseudotensor and $\epsilon_{i_1i_2\ldots i_n}$ has the same definition as its contravariant version). The expression on the l.h.s. can be rewritten as, $$ n^n ϵ^{j_1j_2\ldots j_n} ϵ_{j_1j_2\ldots j_n} $$ We would like to convert the expression on the r.h.s. to the same tensor product. Note that the r.h.s. contains $n$ products of the matrix and its transpose: $$ {A^{j_1}}_{i_1}{A_{j_1}}^{i_1'}, {A^{j_2}}_{i_2}{A_{j_2}}^{i_2'}, \ldots, {A^{j_n}}_{i_n}{A_{j_n}}^{i_n'} $$ If ${A^{j_k}}_{i_k}{A_{j_k}}^{i_k'}= n \delta^{i_k'}_{i_k}$ then, $$ \epsilon^{i_1i_2\ldots i_n}\epsilon_{i_1'i_2'\ldots i_n'} {A^{j_1}}_{i_1}{A_{j_1}}^{i_1'} {A^{j_2}}_{i_2}{A_{j_2}}^{i_2'} \ldots {A^{j_n}}_{i_n}{A_{j_n}}^{i_n'} = \epsilon^{i_1i_2\ldots i_n}\epsilon_{i_1'i_2'\ldots i_n'} n \delta^{i_1'}_{i_1}n \delta^{i_2'}_{i_2} \ldots n \delta^{i_n'}_{i_n} = n^n \epsilon^{i_1i_2\ldots i_n} \epsilon_{i_1i_2\ldots i_n} $$ which since $i_1,i_2,\ldots,i_n$ are dummy indices can be rewritten as, $$ n^n ϵ^{j_1j_2\ldots j_n} ϵ_{j_1j_2\ldots j_n} $$ and is identical to the l.h.s. The definition of a Hadamard matrix is that, $$ AA^T=n I $$ with $\det(A)=n^{\frac{n}{2}}$ and $|{A^j}_i|\leq 1$. Since $\det(A)=n^{\frac{n}{2}} \Rightarrow AA^T=n I$ we have shown that the 'only if' part of the condition holds.
This follows directly from the usual proof of Hadamard’s determinant inequality.
Let $H=QR$ be a QR decomposition, where $Q$ is orthogonal and $R$ is upper triangular. Since $R$ is triangular, the modulus of its $j$-th diagonal element $r_{jj}$ is bounded above by $\|r_{\ast j}\|_2$, the Euclidean norm of the $j$-th column. As $Q$ is orthogonal, we also have $\|r_{\ast j}\|_2=\|Qr_{\ast j}\|_2=\|h_{\ast j}\|_2$. It follows that $$ |\det H|=|\det R| =\prod_j|r_{jj}| \le\prod_j\|r_{\ast j}\|_2 =\prod_j\|h_{\ast j}\|_2 \le\prod_j\sqrt{n} =n^{n/2}. $$ Tie occurs in the first inquality above if and only if $|r_{jj}|=\|r_{\ast j}\|_2$ for every $j$, i.e., iff $R$ is a diagonal matrix. Tie occurs in the second inequality iff $\|r_{\ast j}\|_2=\|h_{\ast j}\|_2=\sqrt{n}$ for each $j$, i.e., iff all elements of $H$ are equal to $\pm1$. Consequently, $|\det H|=n^{n/2}$ iff $R$ is a diagonal matrix whose diagonal elements are all equal to $\sqrt{n}$. This means precisely that $H$ is $\sqrt{n}$ times an orthogonal matrix whose elements are all equal to $\pm1$, i.e., $H$ is a Hadamard matrix.
