# Graph automorphisms and the diamond graph

I thought I understood graph automorphisms, but looking at a specific example is tripping me up. Consider the diamond graph: Apparently, the Klein 4-group is the automorphism group of this graph. The Klein 4-group is $$\{(), (12)(34), (13)(24), (14)(23)\}$$

So, if we take $$\pi=(12)(34)$$ as an example, the definition of graph isomorphism says that $$ij$$ is an edge in $$G$$ if and only if $$\pi(i)\pi(j)$$ is an edge in $$G$$. But if we look at nodes $$2,4$$, then $$24$$ is an edge in $$G$$ but $$\pi(2)\pi(4) = 13$$ is not an edge in $$G$$. So why is $$(12)(34)$$ an automorphism for this graph?

What am I doing wrong? What is the best way to visualize the action of a permutation of the vertices of a graph?

The Klein $$4$$-group is the name given to any group with $$4$$ elements with each element being self-inverse. The graph automorphism group of a graph on $$4$$ vertices is, as you wrote, a subgroup of the symmetric group on $$4$$ elements. The graph automorphism group of the diamond graph you drew is $$\{(), (13), (24), (13)(24)\}$$
This group is a Klein $$4$$-group, just not the one that you wrote.
Hint: Consider $$\{(), (24), (13), (13)(24)\}$$.