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Take $X=S^1\vee S^1$, the wedge of two circles, and label its oriented loops $a$ and $b$. For a fixed $n\ge 3$, letbe permutations of the fibre $\{1,\dots ,n\}$. Construct a labelled graph ${\hat{X}}_n$ with these $n$ vertices, an oriented $a$-edge from $i$ to $\alpha (i)$, and an oriented $b$-edge from $i$ to $\beta (i)$. Map every vertex to the wedge point and each labelled edge homeomorphically onto its corresponding loop. Every vertex has exactly one incoming and one outgoing edge of each label, so this is an $n$-sheeted permutation covering of a wedge of circlesIt is connected because the $n$-cycle $\alpha$ acts transitively on the fibre.

A deck transformation induces a permutation $\gamma$ of the vertices that commutes with both monodromy permutations $\alpha$ and $\beta$. Conversely, such a permutation determines a deck transformation, by the deck transformation group as a monodromy centralizer. The permutations $\alpha$ and $\beta$ generate $S_n$: the conjugates ${\alpha }^j\beta {\alpha }^{-j}$ include the adjacent transpositions, which generate the symmetric group. Hence $\gamma$ centralizes $S_n$ and lies in its center of a group. For $n\ge 3$, the center of $S_n$ is trivial, so $\gamma =1$. A deck transformation fixing one point is the identity by part (a), and therefore