I couldn't find any example of cover which is not Galois.
The definition is following: A cover $p:Y\rightarrow X$ is said to be $Galois$ if $Y$ is connected and the induced map $\bar{p}:Aut(Y|X)\setminus Y\rightarrow X$ is a homeomorphism.
I tried to make some example by using cover of graph $S_1 \vee S_1$, but it didn't work.
Is there simple answer about my question?
Forgive my paint art.
This is a degree 3 cover of $S^1\vee S^1$ (below). The projection map $p$ every red segment or loop to the red loop respecting the given orientation, and similarly for green. You can see it is indeed a cover. Now, I claim that this cover is non Galois. In fact, the deck-automorphism group is trivial. To see this, note that it acts on the black dots. But these black dots are very different from each other : the top one has a single green loop and two red segments out of him, the middle one has two red segments and two green segments, and the bottom one has a single red loop and two green segments. Thus every deck automorphism should preserve these three dots. It follows that every deck automorphism is the identity.