(In this representation of $\phi$, the first row specifies the edges and the second row specifies the two vertices of that edge)
I tried and actually draw some example graphs for $\phi$ and I can conclude that the graph is NOT complete. If it was I could use the theorem that says that a complete graph has $\frac{(n-1)!}{2}$ Hamiltonian cycles.
What can I do now since it's not a complete graph?
If you draw the graph you see the following picture
Recall that a Hamiltonian cycle is a cycle that visits each vertex. In this case there are 4 ways to do this:
We start at vertex 1 and move to vertex 2 with two edges to choose from: a and b
from vertex 2 there is only one way to get to vertex 3: d
from vertex 3 there are two ways to get to vertex 4: e and f
from vertex 4 there is just one way to get to vertex 1: c
Thus we have 4 paths which depend on the choice we make between a and b and between e and f:
$$ adec,\quad adfc,\quad bdec,\quad bdfc. $$
The only possible issue is to make sure that $1 \leftrightarrow 2 \leftrightarrow 3 \leftrightarrow 4 \leftrightarrow 1$ is the only cyclic order we can visit the vertices in. This should be clear from the picture.