Orthogonal orthomorphisms of order 2 EDIT: There is an additional requirement that the composition of the orthomorphisms will also be order 2 (see my answer below).
A full proof is not needed, I will be happy with any argument which convinces me that the statement is true/false.
The motivation came after reading this article: 
http://cms.math.ca/cjm/v13/cjm1961v13.0356-0372.pdf.
It is quite old so I hope someone here knows what advances have been made in this area. I will give the definitions to make the question more self-contained.
Definition: For a group $G$, a permutation $\phi:G\to G$ is an orthomorphism
if $\beta:G\to G$, defined $\beta(x):=x*\phi(x)^{-1}$, is also a permutation. Two orthomorphisms $\phi,\psi$ are orthogonal if $\phi^{-1}\circ\psi$ is an orthomorphism. I say that an orthomorphism $\phi$ is order $n$ if $\phi^n=id$ and $\phi^i$ is an orthomorphism for every $1\leq i<n$.
An orthomorphism is not necessarily an automorphism of $G$, but generally it is assumed that the identity is mapped to itself. My question is:


*

*Is the following statement correct?



For a finite group $G$, there do not exist a pair of orthogonal orthomorphisms of order 2.

If we restrict to orthomorphic automorphisms this is true since only 1 orthomorphic automorphism of order 2 can exist: For any $x\in G$, $\phi(\beta(x))=\phi(x*\phi(x)^{-1})=\phi(x)*x^{-1}=(x*\phi(x)^{-1})^{-1}=\beta(x)^{-1}$. Since $\beta$ is a permutation, this implies $\phi(x)=x^{-1}$.
When dealing with general orthomorphisms, there can exist several orthomorphisms of order 2, but I could not find an example with 2 orthogonal ones (though computation blows up pretty quickly so this doesn't really say much).
Is it possible to use the above restriction of orthomorphic autmorphisms to get a similar restriction for general orthomorphisms?
 A: In canonical form, there is in fact only 1 orthomorphism of order 3, which can't be orthogonal to itself.
As for which groups admit orthomorphisms, that was solved a couple of years ago, the problem was known as the 'Hall-Paige Conjecture', "A finite group with a trivial or noncyclic Sylow 2-subgroup is admissible", meaning that any finite group with a trivial or noncyclic Sylow 2-subgroup will admit orthomorphisms (or complete mappings).  This was posed in 1955.  There has been a considerable amount of work done on this, and was reduced to sporadic simple groups fairly recently.
SPOILER ALERT:
The 'Hall-Paige Conjecture', was finally proven in 2009 (with the exception of $J_4$, after a long road that essentially constructed orthomorphisms for all possible groups, the last being Sporadic simple groups which were mostly handled by Anthony Evans.
Hope this helps!
A: Actually, I forgot to add the requirement that $\phi^{-1}\circ\psi$ is also order 2. So as stated, the statement is false, because the group $C_{13}$ has 2 orthogonal orthomorphisms of order 2. I don't know if anyone other than me is interested in this, but I will give the 2 orthomorphisms just in case.
Anyways, the bounty is still open for the statement with the additional requirement. Hopefully this will make it much easier to prove the statement (probably easier if it is now true :) ).
Denote $C_{13}=\{0,1,\dots,12\}$.
The two orthomorphisms are:
$\phi=(1,2)(3,6)(4,11)(5,9)(7,12)(8,10)$, i.e., $\phi(1)=2$, etc.
it is an orthomorphism since $\beta_{\phi}=id-\phi=(1,12,5,9,4,6,3,10,2)(7,8,11)$.
$\psi=(1,6)(2,9)(3,5)(4,7)(8,12)(10,11)$
and again, easy to check $\beta_{\psi}=id-\psi=(1,8,9,7,3,11)(2,6,5)(4,10,12)$.
and we have
$\phi^{-1}\circ\psi=\phi\circ\psi=(1,9,3)(2,6,5)(4,10,12)(7,8,11),$
which is also an orthomorphism, but unfortunately of order 3 :(
$\beta_{\phi^{-1}\circ\psi}=id-\phi^{-1}\circ\psi=(1,5,3,2,9,6)(4,7,12,8,10,11)$
