Given the iso-regular graph below, I am interested in finding the automorphism of the graph given below.$$AUT(X) = \{\sigma \mid X^{\sigma} = X\}$$
To me it appears that all possible bijections are going to be there so, $n!$ many elements will be there.
On
Here is another way to see that the group is regular on vertices. If you fix a vertex, you fix the unique vertex that is opposite it on the 4-cycle, and the unique vertex that is adjacent to it by an edge not on a 4-cycle. Now, these two type of relations induce a 2-valent graph on the set of vertices, and one can check that it is a 20-cycle. (1-5-7-13-15-17-19-8-6-9-11-18-20-2-4-10-12-14-16-3-1)
Thus, only the identity fixes any vertex, and the group has order dividing 20.
Now, there is an obvious action of the dihedral group of order $10$, that is visible in the picture and, as Jeremy pointed out, you can also exchange the inner and outer cycle. So the automorphism group must have order 20. (And be isomorphic to $AGL(1,5)$, since it embeds in $S_5$.)
If we shrink each of the squares to a single vertex, we obtain $K_5$, which has $S_5$ as symmetry group. Now "un-shrink", and the position of the vertices within the squares is determined by where their adjacent other square is. But not all of $S_5$ is allowed: Once we fix the position of square $1-2-3-4$, this square only allows $D_8$ symmetries.
In other words, we can move square $1-2-3-4$ to any of the five squares and then apply any of the eight symmmetries of the square to it.