Covering space is path-connected if the action of $\pi_1$ on a (single) fiber is transitive Let $p\colon X\to Y$ be a covering map. Suppose that $Y$ is path-connected, locally path-connected and semi-locally simply connected. Let $x,x'\in X$ be two points of $X$. 

$\textbf{Question:}$Is it true that $\pi_1(Y,p(x))$ acts transitively on $p^{-1}(p(x))$ if and only if $\pi_1(Y,p(x'))$ acts transitively on $p^{-1}(p(x'))$?
  Equivalently, is $X$ path-connected if there exists some $x\in X$ such that $\pi_1(Y,p(x))$ acts transitively on $p^{-1}(p(x))$?

Here is my try: Suppose $\pi_1(Y,p(x))$ acts transitively on $p^{-1}(p(x))$. Let $z'\in p^{-1}(p(x'))$. Since $Y$ is path-connected, we may choose a path $\gamma$ from $p(x')$ to $p(x)$. The monodromy functor then induces a map of sets $\phi\colon p^{-1}(p(x'))\to p^{-1}(p(x))$. Let $z=\phi(z')$. Then there exists a loop $\delta\in \pi_1(Y,p(x))$ and a point $\tilde{x}\in p^{-1}(p(x))$ such that the end point of a lift $\tilde{\delta}$ of $\delta$ beginning at $\tilde{x}$ is equal to $z$.
Now what I would like to do is to consider something like the conjugation of $\delta$ with respect to $\gamma$ and lifting $\gamma$ to $\tilde{\gamma}$ and then considering the composition $(\tilde{\gamma})^{-1} \tilde{\delta}\tilde{\gamma}$. However, this does not make sense in general sine we do not know $\tilde{\delta}(0)=\tilde{\gamma}(1)$. Moreover, this idea does not seem to use much of the relation between $z$ and $z'$, $\phi(z')=z$.
$\textbf{Edit:}$ The map $\phi$ can be described more explicitly. I will write that out and see how that helps.  
In the notation from above $\phi(z')=z$ means that there exists a lift $\tilde{\gamma}$ of $\gamma$ starting at $z'$ and ending at $z$. We can do a similar thing with $\tilde{x}$, say $\psi(\tilde{x})=\tilde{z}$ and let $\epsilon$ be the used lifting, where $\psi\colon p^{-1}(p(x)) \to p^{-1}(p(x'))$ is induced by $\gamma^{-1}$ using the monodromy functor. Then $(\tilde{\gamma})^{-1}\tilde{\delta}\epsilon^{-1}=:\tilde{\omega}$ is path from $\tilde{z}$ to $z'$. Letting $\omega=p\tilde{\omega}$ we see that this is a loop at $p(x')$ since $\tilde{z},z' \in p^{-1}(p(x'))$. This proves $[\omega].\tilde{z}=z'$, i.e. $\pi_1(Y,p(x'))$ acts transitively on $p^{-1}(p(x')) $.  
$\textbf{Edit:}$ Actually, this doesn't prove the transitivity. In fact, I had the wrong definition of transitivity in mind when writing the above. However, making a similar approach with the correct definition gives the result. I added an answer in the answer section.
 A: Since my edit in the question does not actually prove that the action is transitive, here is a (hopefully complete) proof. The idea, though, stays the same.  
Suppose that $\pi_1(Y,p(x))$ acts transitively on $p^{-1}(p(x))$. We need to prove that the same is true for $x$ replaced with $x'$, i.e. for each pair $x_1,x_2 \in p^{-1}(p(x'))$ we need to find a loop $\omega$ at $p(x')$ and a lift $\tilde{\omega}$ starting at $x_1$ and ending at $x_2$.  
Fix some path $\gamma\colon p(x')\to p(x)$. The monodromy functor induces a map $\phi\colon p^{-1}(p(x'))\to p^{-1}(p(x))$. Let $z_i=\phi(x_i)$ and let $\tilde{\gamma_i}$ denote the used lifting of $\gamma$, i.e. $\tilde{\gamma_i}(0)=x_i$ and $\tilde{\gamma_i}(1)=z_i$. By assumption, there exists a loop $\delta$ at $p(x)$ and a lifting $\tilde{\delta}\colon z_1\to z_2$. Defining $\tilde{\omega}:=(\tilde{\gamma_2})^{-1}\tilde{\delta}\tilde{\gamma_1}$ (this is a path from $x_1$ to $x_2$) and setting $\omega=p\tilde{\omega}$, the desired result follows.
