The following is from "Homotopical Topology" by Fuchs et al.
Let $p:T \to X$ be a covering, $\overline{x}_0 \in T$, $p(\overline{x}_0) = x_0 \in X$, $D$ the group of deck transformations of $T$, and $N(p_*(\pi_1(T, \overline{x}_0)))$ the normalizer of $\pi_1(T, \overline{x}_0)$ in $\pi_1(X, x_0)$.
Fuchs claims that $N(p_*(\pi_1(T, \overline{x}_0)))/p_*\pi_1(X, x_0) \cong D$ and his proof is as follows:
The orbit of $\overline{x}_0$ under the action of $D$ corresponds exactly to the cosets $\alpha p_*(\pi_1(T, \overline{x}_0)) \in \pi_1(X, x_0)/p_*(\pi_1(T, \overline{x}_0))$ (which have been previously established to be in bijection with the set $p^{-1}(x_0)$) such that $\alpha p_*(\pi_1(T, \overline{x}_0)) \alpha^{-1} = p_*(\pi_1(T, \overline{x}_0))$. QED
What I do not understand is why we can make the conclusion that the points in the orbit correspond exactly to those cosets specified.
My thoughts so far:
Now we also know that there is a deck transformation taking one point $y \in T$ to $y'\in T$ (where $p(y) = p(y')$) iff $p_*(\pi_1(T, y)) = p_*(\pi_1(T, y'))$, and I feel this might have something to do with the the conclusion but I haven't yet discovered the connection.
Any help would be appreciated.
So recall that if our spaces are path connected then the fundamental group is independent of which base point we choose. In particular, if $\alpha: I \to T$ is a path connecting $x_0$ and $x_1$, then we have a map $\varphi: \pi_1(T, x_0) \to \pi_1(T, x_1)$ given by $\gamma \mapsto \alpha^{-1} \gamma \alpha$, and we can see that the groups are conjugate to each other at different base points.
Now recall that the homomorphism of fundamental groups $p_*: \pi_1(T, \overline{x_0}) \to \pi_1(X, x_0)$ induced by the covering $p: T \to X$ satisfies $p_*\pi_1(T, \overline{x_0}) = p_*\pi_1(T, \overline{x_1})$ (where $p(\overline{x_0}) = p(\overline{x_1})$) iff there is a deck transformation taking $\overline{x_0}$ to $\overline{x_1}$. Now since $\pi_1(T, \overline{x_1})$ is conjugate to $\pi_1(T, \overline{x_0})$ this amounts to saying that $p_*(\alpha^{-1})p_*(\pi_1(T, \overline{x_0}))p_*(\alpha) = p_*\pi_1(T, \overline{x_0})$, which is precisely the condition that $p_*(\alpha) \in N(p_*\pi_1(T, \overline{x_0}))$ after combining this with the fact that $p^{-1}(x_0)$ is in bijection with $\pi_1(X, x_0)/(p_*\pi_1(T, \overline{x_0}))$.