Why is $N(p_*(\pi_1(T, \overline{x}_0)))/p_*\pi_1(X, x_0) \cong D$? 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.
 A: 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 $\pi_1(X, x_0)/(p_*\pi_1(T, \overline{x_0}))$.
