Let $k$ be a field, let $G = PGL\_2(k)$ be the projective general linear group of $k$, and let
$X = k \cup \{ \infty \}$ be one-dimensional projective space over $k$. Then $G$ acts on $X$ (via fractional linear transformations). This action has the following properties:


1) The action of $G$ on $X$ is simply 3-transitive. That is, it acts simply transitively on
the set of 3-tuples of distinct elements of $X$. (Edited as indicated in the comments.)


2) Suppose that $x,y \in X$ are distinct elements and that $g \in G$ satisfies
$gx = y$, $gy = x$. Then $g$ has order $2$.


Is the converse true? (That is, if we are given an action of a group $G$ on a set $X$
satisfying 1) and 2), does it follow that $G = PGL\_2(k)$ for some field $k$, 
with its natural action on $k \cup \{ \infty \}$?


(This is true at least when $G$ and $X$ are finite: it can be deduced from the theorem of Frobenius on Frobenius groups.)



Action of PGL(2) on Projective Space
A KT-field $(F,+,\times,\sigma)$
consists of a neardomain $(F,+,\times)$ together with an involutionary
automorphism $\sigma$ satisfying
$$\sigma(1 + \sigma(x)) = 1 - \sigma(1 + x)$$
for all $x \in F \setminus \{0,1\}$. (My impression is that neardomains are quite weak entities, e.g. $F^{\times}$ is required to be a group but it may not be commutative, $(F,+)$ is not even necessarily a group. Industrious MO reader adds the definition of a neardomain to this answer if they wish.)
Sharply $3$-transitive groups are determined up to isomorphism
as permutation groups on $\mathbf{P}^1(F) = F \cup \{ \infty \}$ consisting of maps of the
form:


(i): $x \mapsto a + m x, \quad \infty \mapsto \infty$


(ii): $x \mapsto a + \sigma(b + m x), \quad \infty \mapsto a, \quad - m^{-1} b
\mapsto \infty$,


where $a,b \in F$ and $m \in F^{\times}$. 


Consider the set of elements $\gamma \in G$ such that
$\gamma(0) = \infty$ and $\gamma(\infty) = 0$. They are given exactly
by mappings of the form:
$$\gamma: x \mapsto \sigma( \lambda x)$$
for any $\lambda \in F^{\times}$.
If all such $\gamma$ have order two, then
$$\sigma(\lambda \sigma(\lambda x)) = x$$
 for all $x, \lambda \in F^{\times}$. Setting $x = \lambda^{-1}$, it follows
that $\sigma(\lambda) = \lambda^{-1}$ for all $\lambda \in F^{\times}$.
Since $\sigma$ is an automorphism, it follows that $F^{\times}$ is
commutative. From a Theorem of Kirby (see below), it follows that $(F,+,\times)$
is actually a commutative field, and $G = \mathrm{PGL}\_2(F)$.


All the results and definitions of this answer can be gleamed from
the math review: [MR0997066 (91b:20004a)](http://www.ams.org/mathscinet-getitem?mr=1020218) of a paper by William Kerby



> 
> *A class of canonical sharply 3-transitive groups*, Results Math. **16** (1989), no. 1-2, 89–106 (doi:[10.1007/BF03322647](http://dx.doi.org/10.1007/BF03322647)). 
> 
> 
> 


The paper is only $3$-pages long, so I assume that is is relatively elementary - although I can't access it myself, and it may refer to previous results. (Full disclosure, all I did was type "sharply 3-transitive" into mathscinet,
I don't actually know what a neardomain actually is.)
In case your actual purpose is to generalize this result to $(\infty,\pi)$-whatzit categories with creamy rice pudding centres, you might want to take a glance at the actual paper.


