# If two nodes in a graph are symmetric, does there exist an automorphism that swaps their "location"?

Let's say that there exists a graph G. There exists an automorphism A that maps node n1 onto node n2. Does that imply there exists an automorphism that maps n1 onto n2 WHILE ALSO mapping n2 onto n1?

• What have you tried? Commented Jun 13 at 5:39
• Simple examples. Complete graphs. Trees. Every time I find 2 symmetric nodes, there exists some automorphism that swaps their position. Commented Jun 13 at 5:40

Counterexample: Take the graph with $$12$$ vertices $$A,B,C,D,E,F,G,H,I,J,K,L$$ and $$15$$ edges $$AB,BC,CD,DE,EF,FG,GH,HI,IA,AD,DG,GA,CJ,FK,IL$$. There is an automorphism of order $$3$$ which maps $$J$$ to $$K$$, but no automorphism swaps $$J$$ and $$K$$.
By the way, if we delete the vertex $$L$$, the resulting graph $$\Gamma$$ (with $$11$$ vertices and $$14$$ edges) is an example of something else: the vertices $$J$$ and $$K$$ are not symmetric (the graph $$\Gamma$$ has no nontrivial automorphism) although the subgraphs $$\Gamma-J$$ and $$\Gamma-K$$ are isomorphic.
• +1 For an explicit example. You can save a few vertices by removing $J$, $K$ and $L$ and putting diagonals in the squares instead. Commented Jun 13 at 17:38
Frucht's theorem states that every finite group is the group of symmetries of a finite undirected graph. Take any graph whhose group of automorphisms is, say, the cyclic group of order three, and it will have an automorphism taking some vertex $$u$$ to some vertex $$v$$, and an automorphism taking $$v$$ to $$u$$, but no automorphism taking $$u$$ to $$v$$ and $$v$$ to $$u$$.