Let $p:\cX\to X$ is a covering space of a path-connected, locally path-connected space which admits a universal cover $q:U\to X$, and let $\gamma:x_0\leadsto x_1$ be a curve on $X$. Then there is a map $L_\gamma: p^{-1}(x_1)\to p^{-1}(x_0)$ given by $L_\gamma(\cx_1)=\cx_0$ where $\cx_0=\tilde{\gamma}(0)$ and $\tilde{\gamma}$ is a lift of $\gamma$ ending at $\cx_1$, which is a bijection, since it has an inverse $L_\overline{\gamma}$. If $c_{x_0}$ is the constant loop at $x_0$, then $L_{c_{x_0}}=1_{p^{-1}(x_0)}$, and $L_{\gamma*\gamma'}=L_\gamma L_{\gamma'}$. Furthermore, if $\gamma$ is a loop with endpoint $x_0$, then $L_\gamma\in S_{p^{-1}(x_0)}$. The map $[\gamma]\mapsto L_\gamma$ is then (Since it doesn't depend on the representative of the homotopy class of $\gamma$) an homomorphism; i.e. a permutation representation of an action $\pi_1(X,x_0)$ on $p^{-1}(x_0)$, given naturally by
$$[\gamma]\cdot \cx=L_\gamma (\cx).$$
It can be proved (See Hatcher's Algebraic Topology, pages 68-70) that a covering space is uniquely determined by this action (up to isomorphism). We'll prove a part of it, as an exercise.
Proposition 1. Let now $p:\cX\to X, r:\cY\to X$ be isomorphic covering spaces. Then the group actions induced by them are isomorphic.
Proof. Let $\phi:\cX\to \cY$ be an isomorphism, i.e. a homeomorphism such that $p=r\circ \phi$. Let $\cF=p^{-1}(x_0),\cG=r^{-1}(x_0)$. Then $\phi|_{\cF}$ is a bijection $\cF\to\cG$. Indeed, if $f\in \cF$, then $r(\phi(f))=p(f)=x_0$, so $\phi(f)\in \cG$. Analogously, $\im\phi^{-1}|_{\cG}\subseteq \cF$, so $\phi^{-1}_{\cG}$ is its inverse, and $\phi|_{\cF}$ is a bijection. Let $[\gamma]\in \pi_1(X,x_0)$. We'll prove that $\phi([\gamma]f)=[\gamma]\phi(f)$. Indeed, $[\gamma]f=L^\cX_\gamma(f)=\tilde{\gamma}(0)$ where $\tilde{\gamma}$ is a lift of $\gamma$ in $\cX$, ending at $f$. But $\phi\circ\tilde\gamma$ is a lift of $\gamma$ in $\cY$ ending at $\phi(f)$, thus $\phi(\tilde\gamma(0))=L^{\cY}_\gamma(\phi(f))=[\gamma]\phi(f)$. So $\phi$ is an isomorphism of $\pi_1(X,x_0)-$sets, the required result.
It can be seen, in Hatcher's book, that given a group action of $\pi_1(X,x_0)$ on a set $F$ and a fixed point $u_0$ such that $q(u_0)=x_0$, we can build the covering space $U\times F/\sim$ where $(u,f)\sim (u',f')$ if $q(u)=q(u')$ and $[(q\circ\alpha')(\overline{q\circ\alpha})]f=f'$ and $\alpha:u_0\leadsto u,\alpha':u_0\leadsto u'$. Also, the covering map $p:U\times F/\sim \to X$ is given by $p([u,f])=q(u)$, and the fiber $p^{-1}(x_0)$ is in a 1-1 correspondence with $F$ such that the actions are essentially equivalent.
This said, we can prove the converse of Proposition 1.
Proposition 2. Let $F,F'$ be isomorphic $\pi_1(X,x_0)-$sets. Then the associated covering spaces are isomorphic.
Proof. Take $\phi:F\to F'$ a $\pi_1(X,x_0)-$isomorphism (Which is obviously a homeomorphism, taking discrete topologies), and define the obviously homeomorphism $\varphi:U\times F\to U\times F'$ given by $(u,f)\mapsto (u,\phi(f))$.
See that $(u,f)\sim_F (u',f')$ iff $q(u)=q(u')$ and $[(q\circ \alpha')(\overline{q\circ\alpha})]f=f'$ for $\alpha:u_0\leadsto u, \alpha':u_0\leadsto u'$. But since $\phi$ is a $\pi_1(X,x_0)-$isomorphism, this occurs iff $q(u)=q(u')$ and $[(q\circ \alpha')(\overline{q\circ\alpha})](\phi (f))=\phi (f')$, which occurs iff $(u,\phi(f))\sim_{F'}(u',\phi (f'))$. Thus $\varphi$ becomes a homeomorphism of the quotients:
\begin{CD} U\times F @>{\varphi}>> U\times F'\\ @VVV @VVV\\ U\times F/\sim_F@>{\varphi}>> U\times F'/\sim_{F'}\end{CD}
Furthermore, $p_{F'}\circ\varphi([u,f])=p_F([u,\phi (f)])=q(u)=p_F([u,f])$, so $\varphi$ is really a covering space isomorphism, which is the desired result.
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