Smith theorem implies that the number of Hamiltonian cycles in cubic Hamiltonian graphs is at least two (It also implies that Hamiltonian cubic graphs contain at least three Hamiltonian cycles). Hence, any two Hamiltonian cycles must share some edges. I am interested on lower and upper bounds on the number of shared edges between two Hamiltonian cycles in cubic Hamiltonian graph.
What are the known upper and lower bounds?
Hints. Let $G$ be a cubic Hamiltonian graph of order $n$ and $H_1$, $H_2$ be Hamiltonian cycles of $G$. We denote by the symbol $\gamma(G,H_1,H_2)$ the number of common edges of $H_1$ and $H_2$. The following inequalities hold $$ \frac{n}{2}\leq\gamma(G,H_1,H_2)\leq n-2. $$ Let us call the edges $H_1$ sides of $n$-gon, all other edges of graph $G$ we will call chords.
No path in graph $G$ contains two consecutive chords.
Since each chord splits the remaining vertices of $n$-gon into two parts, each Hamiltonian cycle passing through a chord contains at least two chords.
The left inequality is derived from 1, and the right inequality is derived from 2.
It is easy to construct examples of cubic Hamiltonian graphs of given even order $n$ for which the equality holds.
Examples. In both figures below, $n=8$ and $H_1=123456781$.
On the left $H_2=132456781$ and $\gamma(G,H_1,H_2)=8-2=6$.
On the right $H_2=124378651$ and $\gamma(G,H_1,H_2)=8/2=4$.