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, given two triples $(x,y,z), (x',y',z') \in X^3$, each of pairwise distinct elements of $X$, there is a unique element $g \in G$ such that $gx = x'$, $gy = y'$, $gz = z'$.

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.)

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.)