Could there be closed curves in R^3 with tetrahedral symmetry?

Question

Question

Is there a closed curve in $\mathbb R^3$ that has tetrahedral, octahedral, or icosahedral symmetry? By closed curve I mean a continuously differentiable function $\gamma\colon S^1\to\mathbb R^3$. By tetrahedral symmetry I mean that the symmetry group of the curve contains that of a tetrahedron. Octahedral and icosahedral symmetries are defined similarly.

Motivation

If I want to display a hexagonal prism in a gif animation, it suffices to rotate the prism by 60 degrees and the infinitely looping gif will make it look like the prism is rotating forever.

Now I want to display a tetrahedron $T$ (in general, any polyhedron) in a gif animation. I want to find a way to rotate $T$ so that in the last frame of the gif, the rotated $T$ looks exactly the same as the $T$ in the first frame but they actually differ by a nontrivial rotation.

Answer 1

This is not an answer. Just a way to insert a graphics.

Have you seen this (animated) curve, the 20th image in the Baez document:

Answer 2

One cheating solution is to take the edges of the tetrahedron, duplicate them, and then take a Eulerian Cycle that goes over every edge twice.

If you require non-self-intersection, then I think the symmetry group needs to be the symmetries of a polygon, so it’s not possible. See https://www.ams.org/journals/proc/1982-084-03/S0002-9939-1982-0640242-2/S0002-9939-1982-0640242-2.pdf