Group Action of $Aut(X)$ on edge colored graph

Question

Let $X = (V,E)$ be a graph whose edges are colored $( |V| = n)$ and assume that vertices are numbered from 1 to $n$. Now consider this group action: $$\pi : Aut(X) \times [n] \mapsto [n]$$

I know that $Aut(X)$ preserve the color of the edges i.e. it will map a red edge to red edge only.

I am interested in how orbits going to look in this situation. Are all red edges going to be in a single orbit and all blue edges going to be in a different orbit etc?

I tried on smaller graph, see the diagram below:

In the digram above, Is option 1 correct ?

In general, how the orbits are going to look like ?

EDIT : $Aut(X)$ also preserve the edge colours.

Answer 1

For the non-edge-colored version, the automorphism group is going to be everything, i.e., the symmetric group $$\mathrm{Aut}(X)=\{\mathrm{id},(12),(13),(23),(123),(132)\}.$$ And we can compute there is a single edge orbit $$\{12,13,23\}.$$

Switching to the edge-colored version, we redefine what it means to be an automorphism. Here, automorphisms are not only permutations that maps edges to edges and non-edges to non-edges, but they also preserve edge colors.

By inspection $(13)$ is the only non-trivial permutation of the vertices that maps red edges to red edges and blue edges to blue edges. So we have $$\mathrm{Aut}(X_{\text{edge-col}})=\{\mathrm{id},(13)\}.$$

We can compute that there are two orbits under this group action $$\{13\}$$ since the edge $13$ maps to itself after permuting the vertices according to $\mathrm{id}$ and $(13)$ and $$\{12,23\}$$ since edge $12$ maps to itself after permuting the vertices according to $\mathrm{id}$ and maps to $23$ after permuting the vertices according to $(13)$.

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Specific comments:

• $(12)$ cannot be an automorphism of the edge-colored graph, as the blue edge $13$ maps to the red edge $(23)$.

• Item 2. in the question is not a group, so it can't be the automorphism group of anything.

Answer 2

In general, edges of the same colour will not necessarily be in the same orbit. For instance, consider a square with three edges coloured red and one blue. In this case Aut(X) has three orbits of edges.

In fact, even if you use only one colour, edges don't have to be in the same orbit. It's possible for a graph with $m$ edges ($m \ge 6$) to have no automorphisms at all except the trivial one.