Fix integers $m, n > 0$, where $m$ is the number of sets (or a list if you prefer, this a not a question on set theory) and $n$ is the number of vectors in each set. For each $i=1\ldots m$ let $$x_{i1}, \ldots, x_{in} \in \mathbb{R}^{d_i} \backslash\{0\}$$ be the vectors of the $i$-th set. Additionally, assume that $x_{1r}, \ldots, x_{mr}$ have the same norm, which I denote by $\gamma_r$.
I want to find scalars $a_{ir}$, $i=1 \ldots m, r = 1 \ldots n$ such that
1) for each $a_{ir}$ we have $\| a_{ir} x_{ir} \|^2 \geq \prod_{j \neq i} | \langle a_{js} x_{js}, a_{jt} x_{jt} \rangle |$ for all $s, t$
2) $\prod_{i=1}^m a_{ir} = 1$ for all $r = 1 \ldots n$
I think that something like Lagrange multiplies may work, but I'm not sure. I can't make it work. My hope is that someone here know the way to approach this problem. Thanks.
EDIT: After having a conversation with a friend, he pointed out that we can multiply both sides of 1 by $|a_{is} a_{it}|$ and use 2 to simplify the inequality to $$| a_{ir}^2 a_{is} a_{it}| \| x_{ir} \|^2 \geq \prod_{j \neq i} | \langle x_{js}, x_{jt} \rangle |,$$ so the problem becomes to find the constants such that $$| a_{ir}^2 a_{is} a_{it}| \geq \frac{1}{\| x_{ir} \|^2} \prod_{j \neq i} | \langle x_{js}, x_{jt} \rangle |$$ for all $i, r, s, t$.
EDIT 2: From equation above we may consider the product over $i$, which gives
$$\prod_{i=1}^m| a_{ir}^2 a_{is} a_{it}| \geq \prod_{i=1}^m \left( \frac{1}{\| x_{ir} \|^2} \prod_{j \neq i} | \langle x_{js}, x_{jt} \rangle | \right) \implies$$ $$\implies 1 \geq \prod_{i=1}^m \frac{1}{\| x_{ir} \|^2} \cdot \prod_{i=1}^m \left( \prod_{j \neq i} | \langle x_{js}, x_{jt} \rangle | \right) \implies$$ $$\implies \prod_{i=1}^m \left( \prod_{j \neq i} | \langle x_{js}, x_{jt} \rangle | \right) \leq \prod_{i=1}^m \| x_{ir} \|^2 = \prod_{i=1}^m \gamma_r^2 = \gamma_r^{2m}.$$
This gives new conditions over the vectors. We may assume they are valid to continue the problem.
With the last observation at the second edit, the problem is not interesting anymore. Take the last inequality
$$\gamma_r^{2m} \geq \prod_{i=1}^m \left( \prod_{j \neq i} | \langle x_{js}, x_{jt} \rangle | \right)$$ and consider the case where $r = s = t$. Then we have
$$\gamma_r^{2m} \geq \prod_{i=1}^m \left( \prod_{j \neq i} | \langle x_{jr}, x_{jr} \rangle | \right) = \prod_{i=1}^m \left( \prod_{j \neq i} \| x_{jr} \|^2 \right) = $$ $$ = \prod_{i=1}^m \left( \prod_{j \neq i} \gamma_r^2 \right) = \gamma_r^{2m(m-1)}.$$
From this we conclude that $\gamma_r \leq 1$ for all $r$, so all vectors must be unit length or less. This new restriction is not desired, unfortunately.