Let $X_1\to Y$ and $X_2\to Y$ be two coverings. How does a morphism $X_1\to X_2$ over $Y$ induce morphism $\operatorname{Aut}_Y(X_1)\to \operatorname{Aut}_Y(X_2)$?
It should be trivial, but I can not understand it even in case of $Y$ point: if $X_1\to X_2$ is a mapping of sets, then how does it induce mapping $\operatorname{Aut}(X_1)\to \operatorname{Aut}(X_2)$?
There is no such mapping in general. In your example where $Y$ is a point, you would like to extend $X \mapsto \operatorname{Sym}(X)$, the symmetric group, to a functor $S:\operatorname{Set} \to \operatorname{Group}$. Suppose there is such an $S$. Choose $X_1$ and $X_2$ with $|X_1| = 4$ and $|X_2| = 5$, and choose maps $f: X_1 \to X_2$ and $g: X_2 \to X_1$ with $g \circ f = 1$. The only nontrivial proper normal subgroup of $\operatorname{Sym}(X)$ is the alternating group when $|X| \ge 5$, so there are no onto homomorphisms $\operatorname{Sym}(X_2) \to \operatorname{Sym}(X_1)$. So $S(g)$ is not onto, and $S(g \circ f)$ is not onto. But $S(g \circ f) = S(1_{X_1}) = 1_{S(X_1)}$, so this is impossible. So there is no such $S$.