Is there a simple graph with an odd number of automorphisms (except $1$ and $3$)?

Question

The simple graphs upto $11$ vertices do not have $5,7,9,...$ automorphisms, in other words, the only odd numbers appearing are $1$ and $3$. Is this true for all graphs ?

Formulated as an existence-question :

• Is there an odd number $k\ge 5$ and a simple graph G with $|Aut(G)|=k$?

  In particular, is there a simple graph with $5$ automorphisms ?

Answer 1

I found Frucht's theorem on Wikipedia.
http://en.wikipedia.org/wiki/Frucht%27s_theorem
According to this theorem, any finite group occurs as the automorphism group of a finite, undirected graph.

Answer 2

An old paper of Frucht (referenced on the wikipedia page on graph automorphisms) contains the following results (Theorems 3.1 and 3.2):

If $n>2$, then there is a cubic graph $G$ on $6n$ vertices with $\text{Aut}(G)\simeq \mathbb{Z}_n$.

If $n>3$, then there is a graph $G$ on $3n$ vertices with $\text{Aut}(G)\simeq \mathbb{Z}_n$.

Note also that the proofs explicitly constructs the required graph.