I want to show $Aut(\tilde{X},p)\cong NH/H $ where $H=p_*\pi_1(\tilde{X},\tilde{x_0})$ and $NH$ is the normalizer of $H$.
In my text book, the author sketches the proof of the above by using the facts $Im\phi\cong Im\psi\cong Im\xi=Aut(\tilde{X},p)$. (The group action $g.\tilde{x_0}$ is defined by the terminal point of the lifting of $g=[\gamma]\in NH$ )
I'm stuck at showing the isomorphism between $Im\psi$ and $Im\xi=Aut(\tilde{X},p)$.
Any help will be appreciated.
I am not sure if I understand the proof sketched in your textbook but here's a proof:
$\gamma$ be a loop in $X$ based at $x_0$ which lifts to a path $\tilde{\gamma}$ between $\tilde{x_0}$ and $\tilde{x_1}$ in $\tilde{X}$. $[\gamma]$ belongs to $N(H)$, $H = p_*(\pi_1(\tilde{X}, \tilde{x_0}))$, iff $p_*(\pi_1(\tilde{X}, \tilde{x_0})) = p_*(\pi_1(\tilde{X}, \tilde{x_1}))$. Thus, if $[\gamma]$ is in $N(H)$ then there is a deck transformation $f : \tilde{X} \to \tilde{X}$ such that $f(\tilde{x_0}) = \tilde{x_1}$.
Thus define a homomorphism $g : N(H) \to \text{Aut}(\tilde{X})$ by sending $[\gamma]$ to $f$.
This is a homomorphism, as if $[\gamma], [\gamma']$ are two classes in $N(H)$ such that lift $\tilde{\gamma}$ has endpoints $\tilde{x_0}$ and $\tilde{x_1}'$ and corresponds to the deck transformation $f'$, then $\gamma * \gamma'$ would lift to a path $\tilde{\gamma} * f(\tilde{\gamma}')$ between $\tilde{x_0}$ and $f(\tilde{x_1}') = f(f'(\tilde{x_0}))$. Thus, $[\gamma * \gamma']$ is sent to $f \circ f'$ by $g$.
It's easy to see $g$ is surjective, as given a deck transformation $f$ of $\tilde{X}$ taking $\tilde{x_0}$ to $x \in p^{-1}(x_0)$, $g[\sigma] = f$ where $\sigma$ is a path joining $\tilde{x_0}$ and $x$. The kernel of $g$ consists of loops in $X$ based at $x_0$ which lift to loops at $\tilde{x_0}$ in $\tilde{X}$. Such loops form precisely the group $H$.
Hence, the quotient map $\tilde{g} : N(H)/H \to \text{Aut}(\tilde{X})$ is an isomorphism, as desired.