How many different 2-regular graphs are there with 5 vertices? 
How many different 2-regular (simple) graphs are there with 5 vertices?

I just asked a very similar question, and I actually already understand the answer of this question. 
I think there are very much different ways to look at such a question (but maybe I'm wrong) and my head always kind of feel unsatisfied, even if I get the answer. I hope my head get a little more satisfied by actually understanding the different ways to solve this problem. I see this problem now like:

Just by trial and error I understand that the graph shoud look like:
\begin{array}{ccccc}
& & 1 & & 
\\ & ╱ & & ╲ \\
5 & & & & 2 & \\ | & & & & | \\
4 & & — & &3 
\end{array}
And as there are $4!$ cycles of the form $(12345)\in S_5$, there must be $4!/2$ graphs of this form as $(12345)$ and $(54321)$ are different cycles but represent the same graph.

So this is one way to look at this problem, but I'm interested in other ways to think of this question.
 A: First lets observe that any simple graph $G$ with $5$ vertices and $2$-regular must be connected.
Indeed, assuming that $G$ is disconnected, one component must have 2 or less vertices. But then no vertex in this component can have degree higher than  $2-1=1$.
Then $G$ must have a cycle, since it cannot be a tree. This cycle must have length 3,4 or 5. But this cycle cannot be connected to any other vertex, as the vertices of the cycle already have degree $2$. Thus the cycle is a component of $G$, and since $G$ is connected, it means that the cycle is $G$.
Thus $G$ is a cycle graph, hence $G$ is $C_5$.
As unlabeled graph $C_5$ is the only graph. If you look at labeled graphs, you need to label $C_5$ in all possible ways, and your counting is right: if you fix the vertex $\{1\}$ there are $4!$ ways to label the other vertices and you get each graph twice (clockwise and counterclockwise).
P.S. The argument i used in the third paragraph works more generally: If $G$ is connected and $2$-regular then $G$ is some $C_n$.
P.P.S. If $G$ is $2$-regular but not necessarily connected, then each component is $2$ regular and connected thus some $C_k$. This, together with the observation that $C_k$ must have at least 3 vertices can be used to clasify all simple $2$ regular graphs with $n$ vertices, for small $n$. 
A: There is indeed only one $5$-vertex $2$-regular graph, which is isomorphic to the $5$-cycle:

This graph has the dihedral group (which I will denote $D_5$ today) as its automorphism group.
Let $A$ be the set of labelled graphs isomorphic to the $5$-cycle.  The symmetric group $S_5$ acts on $A$ by permuting the vertex labels.  There is only one orbit under this group action, and this orbit is $A$.  The elements of $S_5$ that stabilise a labelled graph in $A$ are precisely the elements of $D_5$.  Hence, by the Orbit-Stabiliser Theorem $$|A|=\frac{|S_5|}{|D_5|}=\frac{5!}{10}.$$
In general, the same argument can be used to show that the number of labelled graphs isomorphic to a graph $G$ is $$\frac{|V(G)|!}{|\mathrm{Aut}(G)|}$$ where $V(G)$ is the set of vertices of $G$ and $\mathrm{Aut}(G)$ is the automorphism group of $G$.
