Let $U \in \mathbb{R}^n$ be a unitary matrix and $K \subset \mathbb{R}^n$ and $n$-dimensional cube centered at zero. Let \begin{align} K_1=U \cdot K= \{ y: y=U x, x\in K\}. \end{align}
That is $K_1$ is a rotation of $K$. Next, let ${\rm Vol}(\cdot)$ be a volume operator (Lebesgue measure).
Can we give lower bounds on \begin{align} {\rm Vol}( K_1 \cap K), \end{align}
The upper bounds are simple \begin{align} {\rm Vol}( K_1 \cap K) \le \min ({\rm Vol}( K_1 ),{\rm Vol}( K)), \end{align} and tight if $U$ is an identiy matrix.
Lower Bound Based on inclusion Let $B$ be a ball of radius $r$ such that \begin{align} B \subset K_1\cap K, \end{align} then \begin{align} \frac{\pi^{n/2}}{\Gamma(n/2+1)} r^n =Vol(B) \le {\rm Vol}( K_1 \cap K)= 2^nr^n . \end{align}
But, since $\frac{\pi^{n/2}}{\Gamma(n/2+1)} \to 0$ as $n \to \infty$ and $2^n \to \infty$ as $n \to \infty$, the lower and the upper bounds do not match.
Question: Can we come up with a lower bound and upper bounds of the same order as $n \to \infty$.
Consider the $\|\cdot\|_{\infty}$-norm ball of radius $\frac{1}{2},$ $K=\{x\in\mathbb{R}^{n}:\max_{i}|x_{i}|\leq 1/2\}.$ This is a unit cube centered at $0$, and it clearly contains the Euclidean unit ball of radius $\frac{1}{2}.$ If we apply a unitary transformation to $K,$ we get another unit cube centered at $0,$ but the Euclidean unit ball of radius $\frac{1}{2}$ remains unchanged, and is still clearly contained within $UK.$ Thus, given a collection of unitary matrices $\{U_{k}\}_{k\geq 1},$ we actually have that $$V_{n}\left(\frac{1}{2}\right)\leq \mathrm{Vol}\left(\bigcap_{k\geq1} U_{k}K\right),$$ if $K$ is an $n$-dimensional unit cube centered at $0,$ and where $V_{n}(r)$ is the volume of the $n$-dimensional Euclidean ball of radius $r$.