It is a well know fact that if $p:\widetilde{X}\to X $ is the universal covering of $ X$ then $\pi(X,x_0)\cong G(\widetilde{X}) $ where $ G(\widetilde{X})$ is the group of deck transformation of $\widetilde{X}$. The explicit isomorphism is the following:
Fix $x \in p^{-1}(x_0)$. For a given path $ \gamma \in \pi(X,x_0)$ let $\gamma_{x} $ denote the unique lifting of $ \gamma$ starting at $ x$. It is easy to see using the lifting criterion and the fact that $ \widetilde{X}$ is simply connected, that thre is a unique deck transformation $ \tau^x_\gamma$ joining $ x$ and $ \gamma_x(1)$. So $\gamma \mapsto \tau^x_{\gamma} $ is an isomorphism.
It is obvious from the previous that the isomorphism depend on the choice on $ x\in p^{-1}(x_0)$. My question is: ¿which condition is needed to guarantee $\tau^{x_1}_\gamma=\tau^{x_2}_\gamma $ for every $ \gamma$? Here, off course, $ x_1,x_2\in p^{-1}(x_0)$.