Warning: this is a self-answer with major spoilers for one interpretation of the question (all symmetries) - it's different if you don't allow reflections because you can't reflect a piece of paper.
If we use all symmetries, there are
only 5 circuits
Approach:
When we try to look for different ways things might be structured we want to look for invariants - properties of a circuit that don't change when we rotate/reverse/reflect. There are a few that we can take advantage of here. One is that we have three degree-4 nodes (a, b, c in the diagram) and three degree-2 nodes (A, B, C). You can't go from a degree-2 node to another one, so they have to split the 6 times you visit a degree-4 node.
Another is that when you approach and leave a degree-4 node you can have a circuit which intersects itself (if you do A-c-a + B-c-b, for instance) or two ways to not intersect yourself (A-c-B + a-c-b vs A-c-b + a-c-B).
Solution:
Every circuit is equivalent to one of:
1. 444424242 (visiting 4 degree-4 nodes in sequence), non-self-intersecting.
Representative: acbaCbAcB
2. 444424242 (visiting 4 degree-4 nodes in sequence), self-intersecting.
Representative: bCaBcbacA
3. 444244242 (one sequence of 3 degree-4 nodes). This has to approach each of the degree-4 nodes in a different one of the three ways mentioned above.
Representative: aBcabAcbC
4. 442442442 (three sequences of 2 degree-4 nodes). Self-intersecting three times.
Representative: BabAcaCbc* I missed this one in my pen-and-paper enumeration and only noticed after computationally checking
5. 442442442 (three sequences of 2 degree-4 nodes). Self-intersecting twice.
Representative: aBcbCabAc
Illustrations: Link (Please note that upper/lowercase are inverted in the diagram at the top of the image.)
Computational verification:
Colab worksheet with demos