Regular can of course be defined free and transitive. For this post, I choose the definition of regular to be second definition of regular given in Wikipedia.
Let $X$ be a set, possibly empty. Let $G$ be a group, possibly a singleton. Suppose there exists a right group action $\mu: M \times G \to M$.
Unless I misunderstand the meaning of "for every two" or "for each pair", I think the definitions are as follows:
$\mu$ is defined regular if for all $x,y \in M$, there exists a unique $g \in G$ such that $\mu(x,g)=y$
$\mu$ is defined free if for all $g \in G$, if there exists $x \in M$ such that $\mu(x,g)=x$, then we have that $g=1_G$
$\mu$ is defined transitive if $M$ is non-empty and for all $x,y \in M$, there exists a $g \in G$ such that $\mu(x,g)=y$
In proving regular implies free and transitive, how do we prove $M$ is non-empty?
In proving free and transitive implies regular, where do we use $M$ non-empty?
My guess for both: Every $g \in G$ satisfies the equation $\mu(x,g)=y$ if $x,y \in M = \emptyset$.
Related: Free group actions for singleton group or for empty set
You are right. If $M=\emptyset$, the action is regular and free but by definition it can't be transitive. Just impose the condition that $M\neq\emptyset$ and everything will work.