Find Automorphism Group if the following graph (Picture below) I'm studying for an exam and found the following task concerning Automorphism Groups:
"Find the Automorphism $Aut(G)$ Group of the following Graph $G$."

On the outside, the graph looks like a regular pentagon, which would be an easy task. I figured out, that $Aut(G)$ doesn't contain any rotations since the middle part wouldn't be isomorphic. I think I have found one isomorphism, which results in mirroring $G$ on the central vertical axis. This would be the permutation $(24)(9 10)(56)(78)$. I can't find any other isomorphisms, which is hard to believe for me since this task gave 4 out of 40 points on last years exam.
Can anyone help me here? 

Let $I$ be the identity isomorphism, and let $M$ be the mirroring isomorphism. We want to determine whether there exist any other isomorphisms.
In order to do this, let's imagine that there exists an isomorphism that I'll denote $F_1$ which is unequal to either $I$ or $M$. Let's see what we can learn about $F_1$; maybe we will be able to prove a contradiction, or maybe we will be able to find a construction of $F_1$.
An isomorphism must respect the valences of vertices. There are exactly two vertices of valence $2$, namely vertex 5 and vertex 6, and therefore $F_1$ either fixes 5 and 6, or $F_1$ interchanges them. Define a new automorphism $F_2$: 


*

*Case 1: if $F_1$ interchanges $5$ and $6$ then $F_2 = M F_1$ fixes $5$ and $6$; 

*Case 2: if $F_1$ fixes $5$ and $6$ then $F_2 = F_1$. 


Either way, we have produced an isomorphism $F_2$ that fixes $5$ and $6$. 
Furthermore, $F_2$ is different from $I$ (because in Case 1, if $F_2=I$ then $M_1 F_1 = I$ and so $F_1 = M_1^{-1}=M_1$; or if $F_1=F_2=I$; and in either case we get a contradiction).
Next, I look in the graph for all paths between $5$ and $6$. There is just one path of length $3$, namely 5---8---7---6. An isomorphism must take paths to paths and must respect lengths of paths. Therefore $F_2$ take this path to itself and it also fixes the endpoints, and it follows that $F_2$ fixes $7$ and $8$.
Since $F_2$ fixes $5$ it must either fix both $2$ and $8$ or interchange them, but we've just shown it fixes $8$ so it also fixes $2$. By a similar argument $F_2$ fixes $4$.
I think you should probably be able to continue from here, showing that $F_2$ fixes $1$ and $9$ and $10$, and therefore $F_2 = I$. This is a contradiction, and therefore $F_1$ did not exist.
