Does there exist an irreducible polynomial $f \in \mathbb{R}[x, y, z]$ such that:

$$ V := \{ (x, y, z) \in \mathbb{R}^3 : f(x, y, z) \leq 0 \} $$

is a solid of constant width with a finite symmetry group?

The analogous result is true in two dimensions, with an explicit degree-$8$ example given in this paper. If we revolve this curve about its axis of symmetry (or equivalently replace every instance of $y^2$ with $y^2 + z^2$), we would obtain a degree-$8$ surface of constant width depicted below:

This has an infinite symmetry group isomorphic to $O(1)$, so does not answer the question. Similarly, the sphere:

$$ f(x, y, z) = x^2 + y^2 + z^2 - 1 \leq 0 $$

has symmetry group $O(3)$, which is again infinite. Does there exist an algebraic solid of constant width and finite automorphism group?