In an article (paper), there is a description of an outer automorphism of $S_6$. There are six pentagons, arranged with a rule, with vertices $1,2,3,4,5$. Any permutation of these vertices will permute the six pentagons, hence giving a map (homomorphism) from $S_5$ into $S_6$.
I understood this map as follows: if we interchange the vertices, the pentagons will be permuted. Consider permutation $(2 \,3)$, and its effect on the first pentagone a in the note.
(1) In a, vertices $2,3$ are not joined by a "continuous edge", hence after permuting $2,3$ there should not be continuous edge between $2,3$. Hence image of a after this permutation will be either a,d or f (am I correct?). Also, 5 is joined to both $2,3$ before permutation, hence after permuting $2,3$, vertex $5$ should be adjcent to them. We can conclude that a is mapped to a by permutation $(2\,3)$.
(2) The other way, in a, $2,3$ are not joined by continuous edge before permutation $(2\,3)$. Hence after permuting $2,3$, a would be mapped into either a,d or f. Also, $4$ is joined to $2$ but not $3$ before permuting $2,3$; after permuting $2,3$, the vertex $4$ will be joined to $3$ but not $2$; hence image of a by permutation $(2\,3)$ should be d.
I couldn't find my mistake in understanding, if any.
Can one explain the action of $S_5$ on the six pentagons in the article?