Group structure of rotating-square puzzle Suppose we arrange the numbers $1$ through $6$ at the "vertices" of the shape formed by aligning the sides of two squares, as shown below:

In this "puzzle," the only moves allowed are rotating the vertices of either square counterclockwise.
I would like to find the group $G$ that represents this puzzle, but I can't figure out how to account for the interaction between the two squares. All I know right now is that $G\subset S_6$, and that $G$ is generated by the permutations $(1254)$ and $(2365)$.
However, I can't figure out how to express $G$ using well-known groups like $S_n$, $A_n$, $D_n$, and $\mathbb Z_n$, the direct product $\times$, and the semidirect product $\rtimes$ (with no corresponding homomorphism specified).
Can someone please show me how to find the group corresponding to this game?
NOTE: To someone who is experienced with group theory, this is probably an easy exercise; however, to a novice like myself, this is quite confusing

GAP shows that the group is in fact isomorphic to $S_5$. A geometric interpretation was requested for what $5$ things are being permuted. Consider the following $5$ sets of edges between the vertices.
$(1,2,5,4)\leftrightarrow(orange,blue,purple,green)$
$(2,3,6,5)\leftrightarrow(red,blue,purple,green)$

To see that this is all of $S_5$ and not some subgroup, compute some products of elements.
$(orange,blue,purple,green)*(red,blue,purple,green)=(red,purple,orange,blue,green)$ 
which has order $5$, so the order of the group is a multiple of $5$.
$(orange,blue,purple,green)*(red,blue,purple,green)^{-1}=(red,orange,blue)$ 
which has order $3$, so the order of the group is a multiple of $3$.
And $(orange,blue,purple,green)$ has order $4$ so the order of the group is a multiple of $4$.
Now the order of the group must be a multiple of $3*4*5=60$, so either $S_5$ or $A_5$. But we have elements of order $4$ in our group, which leaves only $S_5$.
