Do there exist asymmetric hamiltonian graphs? A graph is asymmetric if its automorphism group is trivial. I have tried to create some (very small) hamiltonian graphs but it seems non trivial to force them to be asymmetric. 

Yes, there do. Here is an example. Arrange vertices numbered 1 to 7 cyclically, so that 7 connects back to 1. Now add edges (1, 3), (1, 4), (2, 5), (2, 6), (2, 7).
Proof that the automorphism group is trivial: Vertex 1 is the only vertex of degree 4, and vertex 2 is the only vertex of degree 5, so these vertices are fixed by any automorphism. Vertex 4 is also fixed by the condition of being adjacent to 1 but not to 2. Vertex 3 is uniquely adjacent to 1 and 4, so it is fixed. Vertex 5 is the only one left that is adjacent to 4, so it is fixed, and similar arguments apply to 6 and 7.
