Proving that $|\operatorname{Aut}(C_n)|=2n$. 
Question: Prove that $|\operatorname{Aut}(C_n)|=2n$.

This the question $2$ chapter $2$, in the book Algebraic Graph Theory, Godsil and Royle.
I've managed to prove that $\left|\operatorname{Aut}(C_n)\right|\geq2n$ considering that the subgroup generated by the permutation $(12\ldots n)$ and the reflexion.
I haven't been able to prove equality.
 A: Let $v_1, v_2, \dots, v_n$ be the vertices of $C_n$, with edges $v_i v_{i+1}$ for $i=1, \dots, n-1$ as well as $v_1 v_n$.
To pick an automorphism of $C_n$, we need to:

*

*Pick $k \in \{1,2,\dots,n\}$ and decide that $v_1 \mapsto v_k$. Of course, one of these possibilities has to be true, and there's $n$ of them.

*Because $v_1 v_2$ has to map to another edge, we know that $v_2 \mapsto v_{k-1}$ or $v_2 \mapsto v_{k+1}$. Choose one of these options, and already you have $2n$ possibilities.

So now it remains to show that each of these choices can be completed to only one automorphism of $C_n$. (You've already shown that each choice can be completed to at least one automorphism.) To do this, check that there's only one possibility for $v_3$, and then that leaves only one possibility for $v_4$, and so on by induction until you get to $v_n$.
(Note that in general this strategy also requires us to check that at the end, what we've defined really is an automorphism. We can skip that here, because you've already proved that at least $2n$ automorphisms exist, and here we see that there are at most $2n$.)
