I want to show that for a connected covering $p:E\to X$, if $\phi\in \text{Aut}(E,p)$ is a covering automorphism, then for every evenly covered open set $V\subseteq X$ s.t. $p^{-1}(V)=\bigsqcup_{\alpha}U_\alpha$ we have that $\phi$ permutes the open sets $U_\alpha$. I know it permutes the points in the fibers, and I know that $p(\phi(U_\alpha))=V$ but what if $V$ is not connected? I'm thinking about the fact that a part of $U_\alpha$ can be mapped in some $U_\beta$ and another one in some $U_\gamma$. Why can't this happen?
I managed to prove it in the case $X$ is locally connected: in that case I can always find connected evenly covered neighborhoods for points in $X$. So if $V$ is connected and $p^{-1}(V)=\bigsqcup_{\alpha}U_\alpha$, then $\phi(U_\alpha)\subseteq U_\beta$ for some $\beta\neq \alpha$ (since the action of $\text{Aut}(E,p)$ is properly discontinous) and then we cannot have $\phi(U_\alpha)\neq U_\beta$ since $p(\phi(U_\alpha))=V$ and $p$ is injective if restricted to $U_\beta$.
Sure, consider $S^1=[0,1]/(\{0\}\sim\{1\})$ covered by $\Bbb R$.
Let $V=V_1\cup V_2$ be the union of two open disjoint intervals in $(0,1)$. You can write $$p^{-1}(V)=\bigcup_n (a(n)+V_1)\cup(b(n)+V_2)$$ where the addition of a number in $\Bbb R$ and a subset of $\Bbb R$ is defined in the obvious way. $a$ and $b$ are any bijections from $\Bbb Z\to\Bbb Z$.
If you have $\phi$ be the translation by $1$ then $$\phi\big((a(n)+ V_1)\cup( b(n)+V_2)\big)=\big((a(n)+1)+V_1\big)\cup\big((b(n)+1)+V_2\big)$$ now $a(n)+1=a(m)$ for some $m$, but there is no need for $b(n)+1=b(m)$ for the same $m$. So $\phi$ does not permute the leaves, because you can shuffle the components of the leaves around as you see fit.