When a covering transformation is a covering? Given a covering $c:\tilde{X}\to X$, is there any set of conditions in order to have $T:\tilde{X}\to \tilde {X}$ to be a covering transformation provided $p=p\circ T$?
I know that if T is the identity, the problem is trivial. But is it true for any other T that is NOT the identity?
 A: For $T$ to be a covering transformation all we need is $p=pT$. While it's true that the identity satisfies this, we can have many more possible $T$'s. Take for example the covering map $p : \Bbb{R} \rightarrow S^1$ given by the standard $p(x)=e^{2 \pi i x}$. Then integral translations are valid covering transformations. Namely, take $T: \Bbb{R} \rightarrow \Bbb{R}$ given by $T(x) = x + n$ where $n \in \Bbb{Z}$. Then it's easy to see that $p(x) = pT(x)$. 

If you want $T$ (a covering transformation satisfying $p=pT$) itself to be a covering map: $T$ is always a covering map because $T$ is always a homeomorphism and homeomorphisms are always covering maps.
A: Good question.
You can consider $\Lambda$ the set of all covering of $X$ of total space $X^\sim$ and on this set the group $\hom(X^\sim ,X^\sim)$ acts it because you can prove that if $p\in\Lambda$ and $\psi\in \hom(X^\sim ,X^\sim)$ than $p\circ \psi \in \Lambda$.
So in this perspective your question is to find ${\rm Stab}(p)$ with rispect to this action.
For first think you can observe that Stab($p$) is a subgroup of  $\hom(X^\sim ,X^\sim)$. There is some relationship between this group and $\pi_1(X,p(e_0))$? 
Yes, you can always define a map $\Phi: \pi_1(X,p(e_0))\to{\rm Stab}(p)$ that maps each $\gamma^\sim$ to $\Phi(\gamma^\sim):X^\sim \to X^\sim$ such that $\Phi(\gamma)(x)={\rm lifting}_{x}((p\circ l_x)\gamma \overleftarrow{(p\circ l_x)} )(1)$ where $l_x $ is a path between $x$ and $e_0$ on $X^\sim$
