Theorem 1   A group action is simply transitive if and only if it is free and transitive

Proof. 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.