For which knots $K$ is there an embedding of $K$ in $\mathbb{R}^3$ for which $x\in K\iff (-x)\in K$?
Obviously, the unknot can be: take the circle given by $(\sin(t),\cos(t),0)$ for $t\in[0,2\pi)$.
More generally, given a knot $K$ and its mirror image $K'$, the knot sum $K\# K'$ can be embedded in a centrally symmetric manner by placing $K$ far from the origin, placing its negated copy, and bridging the gap between them by two straight lines.
In the negative direction, any chiral knot can't be embedded symmetrically, because the inversion of $\mathbb{R}^3$ reverses chirality but fixes such an embedding. (I think this can be made more precise with notions of amphichirality, but I'm not sure I understand the definitions and consequences well enough to make confident statements here.)
What other things can we prove about such embeddings? At a minimum, does anyone have an example of a centrally symmetric drawing of a nontrivial prime knot?
Any knot with an involution which reverses the orientation on $\mathbb{R}^3$, but preserves the orientation on the knot has an embedding of the type you describe. (Equivalently an involution which reverses the orientation on the meridian but preserves the orientation on the longitude.)
I believe such knots are usually called "strongly positive amphicheiral" in the literature. See for example Figure 1(c) in this paper: https://projecteuclid.org/download/pdf_1/euclid.rmjm/1250127863
For a specific hyperbolic knot I would use Snappy (https://snappy.math.uic.edu）to check if the symmetry group has such an element. For example, $4_1$ has exactly two symmetries which reverse the orientation on $S^3$ but preserve the orientation on $4_1$; however, both of these symmetries have order 4 (not 2) so that $4_1$ is not strongly positive amphicheiral.
You can see these order 4 symmetries in the following diagram as a $2\pi/4$ rotation (in either direction) followed by reflection across an $S^2$ intersecting the diagram in the dotted green circle.
https://i.stack.imgur.com/AJeSJ.png