$G$ acts on $X$ with property: For any two ordered pairs $(x,y)$ and $(x',y')$ there exist $g\in G$ st. $x'=x^g$ and $y'=y^g$. $G$ acts primitively?

Question

Let $G$ acts on $X$ with the property: For any two ordered pairs $(x,y)$ and $(x',y')$ such that $x$ and $y$ are distinct elements in $X$, as well as $x'$ and $y'$, there exist $g\in G$ such that $x'=x^g$ and $y'=y^g$. Show that $G$ acts primitively on $X$.

We know that $G$ acts primitively on $X$ if:

1. $G$ acts transitively on $X$
2. all blocks are trivial

1. $\forall x,x'\in X, \exists g\in G$ st. $x'=x^g$, so transitivity follows by the definition?

2. Suppose $B$ is block and $|B|>1$. WTS: $B=X$
   $B^g=\{ x^g:x\in B \}=\{ x':x'\in X \}=X$. Similarly for $y$.

So it should be it, our action should be primitive, but somehow I'm sceptical about this. Does this make sense?