Proposition 1: Let ${\cal O}_x$ be a $p\times n$ matrix in the form of (1) or (2) whose entries are real linear combinations of real variables $x_1, x_2, ..., x_k$. $${\cal O}_x=x_1A_1+x_2A_2+\cdots+x_kA_k\tag{1}$$ where $A_1, A_2, ..., A_k$ are constant real matrices in $\mathbb{R}^{p\times n}$. $${\cal O}_x=(B_1x\ \ B_2x\ \ \cdots\ \ B_nx)\tag{2}$$ where $x\in\mathbb{R}^k$ and $B_1, B_2, ..., B_n$ are constant matrices in $\mathbb{R}^{p\times k}$.
Then, prove the following three conditions are equivalent.
1) ${\cal O}_x$ is a real orthogonal design; i.e., it satisfies (3). $${\cal O}_x^{\sf T}{\cal O}_x=(x_1^2+x_2^2+\cdots+x_k^2)I_{n\times n}\tag{3}$$
2) There exist $k$ constant matrices $A_1, A_2, ..., A_k\in\mathbb{R}^{p\times n}$ such that $$A_i^{\sf T} A_i=I_{n\times n}, i=1, 2, ..., k\tag{4a}$$ $$A_i^{\sf T} A_j+A_j^{\sf T} A_i=0_{n\times n}, 1\leq i<j\leq k.\tag{4b}$$
3) There exist $n$ constant matrices $B_1, B_2, ..., B_n\in\mathbb{R}^{p\times k}$ such that $$B_i^{\sf T} B_i=I_{k\times k}, i=1, 2, ..., n\tag{5a}$$ $$B_i^{\sf T} B_j+B_j^{\sf T} B_i=0_{k\times k}, 1\leq i<j\leq n.\tag{5b}$$
This is a proposition in the paper "Orthogonal designs with maximal rates." The author wrote that by using Lemma 1, (1), and (2), we can easily see that a real orthogonal design ${\cal O}_x$ satisfying (3) has an equivalent characterization in terms of matrix equations, as stated in Proposition 1. The author does not provide a proof for Proposition 1. Therefore, I want try to prove it by myself. However, I have no idea how to begin. Any comments or hints? Thank you very much.
Lemma 1: Let $k\in\mathbb{N}, n\in\mathbb{N}, p\in\mathbb{N}, A\in\mathbb{R}^{k\times k}, B\in\mathbb{R}^{k\times n}, C\in\mathbb{R}^{n\times n}, A_i\in\mathbb{R}^{p\times n}$ for $i=1,2,...,k,$ and $B_{jl}\in\mathbb{R}^{p\times n}$ for $j,l=1,2,...,k$ with $j<l$. Then, we have the following.
1) $x^{\sf T}Ax+x^{\sf T}By+y^{\sf T}Cy=0$ for all $x\in\mathbb{R}^k$ and all $y\in\mathbb{R}^n$ if and only if $A^{\sf T}+A=0_{k\times k},B=0_{k\times n},$ and $C^{\sf T}+C=0_{n\times n}.$
2) The equality $$\sum_{i=1}^k x_i^2A_i+\sum_{1\leq j < l \leq k}x_jx_lB_{jl}=0_{p\times n}\tag{6}$$ holds for all $x=(x_1, x_2, ..., x_k)^{\sf T}\in\mathbb{R}^k$ if and only if $A_i=B_{jl}=0_{p\times n}$ for $1 \leq i \leq k$ and $1\leq j < l \leq k$.
Consider this a partial answer, since I will only deal with $\cal O_x$ in the form $(1)$, thus I will limit myself in proving that condition $1$ and $2$ are equivalent. It should not be hard to extend the proof in the case of form $(2)$. As a slight change in notation, I will use $K$ instead of $k$ (since $k$ is usually an index, while $K$ is fixed). As another notation change, I will use $I_n$ instead of $I_{n \times n}$ to denote the identity matrix of order $n$.
Firstly, notice that notice that $x_1, \dots,x_K$ are scalars and rewrite $(3)$ in a more useful way:
$ \begin{align} (3.1) \ \mathcal{O}^T_x \mathcal{O}_x = I_n \sum_{i=1}^K x^2_i \end{align} $
Notice that, for scalars,
$ \begin{align} \left( \sum_{i=1}^n x_i \right)^2 = \sum_{i=1}^n x_i^2 + 2\sum_{1\leq i<j \leq n} x_ix_j \end{align} $
In our case, it should be simple enough to prove that
$ \begin{align} (*) \quad \mathcal{O}^T_x \mathcal{O}_x = \sum_{i=1}^K x^2_i A^T_i A_i + \sum_{1 \leq i < j \leq K}x_i x_j \left(A_j^T A_i + A_i^T A_j\right) \end{align} $
Part 1. condition $2 \Rightarrow$ condition $1$. This can simply be proved by direct calculus (in fact, this is an obvious fact). Enforcing the hypothesis in $(*)$, you have that the quadratic terms $x_i^2$ are multiplied by $A_i^T A_i = I_n$, while the mixed products are zero, since $A_j^T A_i + A_i^T A_j = 0_n$.
Part 2. condition $1 \Rightarrow$ condition $2$. Now there is a problem in the formulation of the proposition. Hurwitz-Radon conditions are usually different from what is written in Proposition 1 (see note). The truth is that condition $1 \not\Rightarrow$ condition 2, because $(3)$, as it is written, holds only for that specific linear combination and this becomes a problem, since we cannot apply Lemma 1, in particular $(6)$, which must hold for all linear combinations.
To fix the ideas, try writing $\mathcal{O}^T_x \mathcal{O}_x$ for $K=2$ and $x_1=0$. You will notice that you loose information about $A_1$ matrix, i.e. you cannot conclude that $A_1^T A_1 = I_n$ (and you also loose information on the mixed products). Nonetheless, condition $2$ is not in contradiction with condition $1$: it does not follow from it, but it's still compatible with it.
The "correct" proposition should have been worded in a different way; more precisely, $(3)$ must hold for all linear combinations. If that is the case, then:
$ \begin{align} (**) \quad I_n \sum_{i=1}^K x^2_i = \sum_{i=1}^K x^2_i A^T_i A_i + \sum_{1 \leq i < j \leq K}x_i x_j \left(A_j^T A_i + A_i^T A_j\right) \quad \forall \mathbf{x} \end{align} $
where $\mathbf{x}$ is a vector which contains all the scalars $x_i$, with $i=1,\dots,K$. Since $(**)$ holds for all linear combinations, applying $(6)$ (or even without it), it is easy enough to show that $A_j^T A_i + A_i^T A_j = 0_n$ and $A_i^T A_i = I_n$, thus closing the proof.
This answer probably needs some refinements and more details, but the sketch of the proof should be good enough for your purpose.
Note: Try looking here.