Why this graph has automorphism group is isomorphic to the cyclic group of order 4?

Question

My professor say that this graph is a non-separable graph whose automorphism group is isomorphic to the cyclic group of order 4 without telling me why

I can see this graph has no cut vertex, so it's non-separable. Is can be rotated and flipped symmetrically in manner $R^2 =I=r=r^2$ where $R$ is flipping vertically or horizontally, and $r$ is rotating the graph counter clockwise by 90 degree.

But I still can't see why this graph has automorphism group is isomorphic to the cyclic group of order 4?

Answer 1

Hint:

Each of automorphism carries a vertex with maximal degree to vertex to maximal degree. Furthermore, there are $4$ vertices with the degree $5$. Hence we have $4$ automorphisms and we can conclude that Aut$(G)\cong \mathbb Z_4$. Note that it is not hard to see that there is an automorphism with order $4$.

Now, assume that $\phi\in Aut(G)$ fixes all vertices with degree $5$. Since each vertex with degree $5$ has a neighborhood whose degree is $2$, we can conclude that $\phi$ fixes all vertices with degree $2$.

Automorphism $\phi$ fixes all vertices with degree $5$ and $2$ and so ....

Answer 2

$|Aut(G)|=4 $ because $Aut(G)$ ={$a$, $a^2$, $a^3$,$a^4$ } where is $a$, 90 degree clockwise rotation of graph. then $Aut(G)$ is cyclic group of order 4. since there 2 group of order 4 then $Aut(G) \cong \mathbb Z_4 $ or $Aut(G) \cong \mathbb Z_2 *\mathbb Z_2 $