The Structure of Antimorphisms of a Self-complementary Graph It is a well-known result proven by Gibbs in 1974 that every self-complementary graph of order $4n$ can be decomposed to $n$ disjoint $P_4$. Sachs and Ringel proved that any antimorphism of a self-complementary graph of order $4n$ is a product of circular permutations, which are called ${\it{cycles}}$ of this antimorphism, whose lengths are all multiples of $4$. Obviously, the subgraphs induced by each cycle are self-complementary. Since $P_4$ is also a self-complementary graph, here comes my question: is there any recent result showing whether each self-complementary graph has an antimorphism, the lengths of whose cycles are all $4$, or not?
 A: After checking in Mathematica, I found that the Paley graph of order $17$ (House of Graphs page) is a counterexample.

It has $136$ antimorphisms, and all of them are $16$-cycles leaving one vertex fixed. This much I got from the FindGraphIsomorphism command in Mathematica, and I don't have a good argument for why these are all the antimorphisms other than "Mathematica said so".
However, we can figure out what all $136$ antimorphisms are from the definition of the Paley graph: when we number its vertices $\{0,1,2,\dots,16\}$, there is an edge $ij$ whenever $i-j$ is a quadratic residue modulo $17$. Therefore:

*

*Multiplication modulo $17$ by a quadratic nonresidue (one of $3, 5, 6, 7, 10, 11, 12, 14$) is an antimorphism of the graph that leaves vertex $0$ fixed.

*We can number the vertices $0, 1, \dots, 16$ starting at any vertex, giving us $8$ antimorphisms that fix each of the $17$ vertices, for a total of $8 \cdot 17 = 136$ antimorphisms.

These antimorphisms are all $16$-cycles because the quadratic nonresidues must all be primitive roots: by Euler's criterion, if $q$ is a quadratic nonresidue modulo $17$, then $q^8 \equiv -1 \pmod{17}$, so $q^{16}$ is the least power of $q$ congruent to $1$ modulo $17$.
The Paley graph does not have order $4n$. However, deleting any vertex from it gives a $16$-vertex self-complementary graph with $8$ antimorphisms that are all $16$-cycles. (These are induced by the $8$ antimorphisms of the Paley graph that fix the vertex we deleted.)

There are smaller examples, too. Checking the $720$ self-complementary vertices of order $12$ from Brendan McKay's page, I found $12$ examples in which every antimorphism is the product of a $4$-cycle and an $8$-cycle. I don't have nice stories to tell about them, but here is the Graph6 code for all $12$ of them:
>>graph6<<K??FEejRxnb~
K??KjRFixnj^
K?KtSlcUK^m^
K?dcjVHhhriv
K?dcrNHhhriv
K@?LShfUs^m^
KSP@OkfE|xnx
K_CeKmjRjefN
K_KtSlcUK^m]
KsP@OkfE|xnw
KwC[tNILYZMR
KwC]dR_Fy^E{

