Consider the finite cross $C$ (=union of line segments $\overline{(0, -1)(0, 1)}$ and $\overline{(-1, 0)(1, 0)}$) and the unit half-circle $H$. It is easy to see that we may pack continuum-many disjoint copies of $H$ into the plane $\mathbb{R}^2$ without overlapping, whereas for $C$ we may never pack uncountably many copies without overlapping. *By "copy," here, I mean "copy up to rotation and translation," or more generally "image under an element of the affine special orthogonal group." The specific choice of transformation group to use isn't important, so long as it is reasonable.*

For $S\subseteq\mathbb{R}^n$, let $\mathfrak{p}(S)$ be the supremum of the cardinalities of disjoint packings of copies of $S$ inside $\mathbb{R}^n$. In case the continuum hypothesis holds, either $\mathfrak{p}(S)$ is countable or $\mathfrak{p}(S)=2^{\aleph_0}$; however, in general this need not be true. For example, let $G\subseteq (\mathbb{R}, +)$ be a subgroup of index $\aleph_1$ and let $H$ be a coset of $G$; then the copies of $H$ are precisely the cosets of $G$, so $\mathfrak{p}(H)=\aleph_1$.

My question is: suppose $\neg CH$. Under what assumptions - both on the shape $S$ and the ambient axioms of set theory - can we conclude that $\mathfrak{p}(S)$ is either countable or continuum? For example:

Does ZFC prove that, for $S\subseteq\mathbb{R}^n$ compact, $\mathfrak{p}(S)$ is either countable or continuum?

I suspect the answer is "yes," but I don't see how to prove it.