Let $B$ denote the unit Euclidean ball centered at $0 \in \mathbb{R}^n$. Given a set $K \subset \mathbb{R}^n$ let us denote by $P_K(x)$ the maximal number of points $y_i \in K \cap (x + B)$ such that the Euclidean distance between $y_i, y_j$ is strictly greater than $1/2$. In other words it measures how many well-separated points we can put locally around $x$ in $K$.
Question: Suppose $K$ is sign-symmetric ($K = -K$), convex, and compact. Is it true that $P_K$ is maximized with $x = 0$?
Intuitively, this should be the case, since (assuming additionally, $K$ has nonempty interior) $K$ has maximal volume near the origin. Indeed, we know that $\mathrm{vol}_n(K \cap (x + B))$ is maximized at $x = 0$. So, intuitively the origin is the most "massive" and therefore should lead to the largest packing numbers.
One easy observation is that for $x \in K$, we can write $y \in K \cap (x + B)$ as $x + 2m$ where $m = (y-x)/2$. On the other hand, clearly $2m \in 2K \cap B$ under our assumptions. In other words $K \cap (x + B) \subset x + 2K \cap B$. This gives the slightly weaker inequality $$ \sup_{x \in K} P_K(x) \leq P_{2K}(0). $$
