Let $\mathfrak X\subset\Bbb R^d$ be a full-dimensional and countable subset. Full-dimensional means that $\mathrm{span}(\mathfrak X)=\Bbb R^d$. How large can its symmetry group
$$\mathrm{Aut}(X):=\{T\in\mathrm O(\Bbb R^d)\mid T\mathfrak X=\mathfrak X\text{ set-wise}\}.$$
be? I suspect it can be at most countable, but I have no argument for that.
It is indeed at most countable. By the full-dimensionality assumption $\mathfrak X$ contains a basis $e_1,\dots, e_d$ of $\mathbb{R}^d$. You may then define the map $$ \begin{array}{ccccc} \Phi & : & \textrm{Aut}(\mathfrak{X}) & \to & \mathfrak{X}^d \\ & & T & \mapsto & (T e_1,\dots, Te_d). \end{array} $$ Since the elements of $\textrm{Aut}(\mathfrak{X})$ are by assumption linear, $\Phi$ is injective. However, since $\mathfrak{X}$ is countable, so is $\mathfrak{X}^d$ and the result follows.