If a function (like a hash function) maps a vertex of a connected Cayley graph to another vertex which will be the ending point of a Hamiltonian path, is there a particular advantage over a function mapping the vertex based on any other shorter path?
The function will always follow a same set of edges in the path. As an example for a function $f$, if the hamiltonian path in the Cayley graph is given by $s\cdot t^2 \cdot s^2 \cdot t \cdot s^3$, then for vertices $v_1, v_2$ of the graph, $f(v_1) = v_1 \cdot s\cdot t^2 \cdot s^2 \cdot t \cdot s^3 =v_2$.
For another two vertices, $v_3,V_4$, $f(v_3) = v_3 \cdot s\cdot t^2 \cdot s^2 \cdot t \cdot s^3 =v_4$ etc. always follow the same set of edges from the starting vertex, which is possible in Cayley graphs since each vertex has the same environment.
If it is any other path like $s \cdot t \cdot s^3 \cdot t$, for a function $g$ and vertices $w_1,w_2$ in the Cayley graph it will be, $g(w_1) = w_1 \cdot s \cdot t \cdot s^3 \cdot t =w_2$, for vertices $w_3,w_4$, $g(w_3) = w_3 \cdot s \cdot t \cdot s^3 \cdot t =w_4$ etc. Again the same set of edges will be traced to reach the destination vertex.
Then is there any advantage in particular, when the path chosen is Hamiltonian path?
As I understand even if a shorter path is chosen, each vertex will get mapped to a unique vertex, and the function will be one to one?