Necessity follows since $g.x=x$ implies that $g=1_{G}$ because $1_{G}.x=x$ also. Now assume the action is free and transitive and we have elements $g_{1},g_{2}\in G$ and $x,y\in X$ such that $g_{1}.x=y$ and $g_{2}.x=y$. Then $g_{1}.x=g_{2}.x\implies g_{2}^{{-1}}.g_{1}.x=(g_{2}^{{-1}}g_{1}).x=x$ hence $g_{2}^{{-1}}g_{1}=1_{G}$ because the action is free. Thus $g_{1}=g_{2}$ and so the action is simply transitive. ∎