(Orbit Criterion) Let $p:\tilde X \to X$ be a covering map. If $\tilde q, \tilde q' \in \tilde X$ are two points in the same fiber $p^{-1}(q)$, there exists a covering transformation taking $\tilde q$ to $\tilde q '$ if and only if the induced subgroups $p_{*}\pi_{1}(\tilde X, \tilde q)$ and $p_{*}\pi_{1}(\tilde X, \tilde q')$ are equal.
It is well known lemma in a covering group theory, however I am wondering about example of some neat covering space with different induced fundamental groups $p_{*}\pi_{1}(\tilde X, \tilde q)$ and $p_{*}\pi_{1}(\tilde X, \tilde q')$ (where $\tilde q, \tilde q '$ are in the same fiber). I used so far only very ordinary covering spaces so I cannot come up with such example and hence my question.
Let $X$ be a circle. Let $p : \widetilde X \to X$ be the universal cover, where $\widetilde X$ is a line. It follows that $X \sqcup \widetilde X$ with the obvious map is a covering space of $X$ that satisfies the property you're looking for. Just pick $\tilde q \in X$, $\tilde q' \in \widetilde X$.
If you want a path-connected covering space, consider the $2$-oriented graph $X$ whose fundamental group is the free group on two generators $\langle a, b\rangle$. The subgroup $\langle a\rangle$ is not a normal subgroup of $\langle a, b\rangle$, and hence the covering space $\widetilde X$ that corresponds to $\langle a\rangle$ is also not normal. Let $q$ be the base point of $X$. Depending on the choice of $\tilde q$, the subgroup $p_*(\pi_1(\widetilde X, \tilde q))$ will have the form $\langle b^i ab^{-i}\rangle$ for some fixed $i$. Here is a visualization of $\widetilde X$ (courtesy of Hatcher's Algebraic Topology):