I am having a hard time showing that every finite group is the automorphism group of some infinite degree covering space of a bouquet of circles (rose). Here's what I have done so far:
Let $G = \{g_1,\ldots,g_n\}$ be a finite group and construct its Cayley graph $\Gamma$ with respect to some finite set of generators (e.g. all the $g_i$'s). Let $B$ be a bouquet of circles where each circle corresponds to a generator. Then we know that there exists a regular covering map $p : \Gamma \to B$ such that $\mathrm{Aut}(\Gamma) \cong G$. We wish to extend $p$ to an infinite degree cover $\tilde{p} : \tilde{\Gamma} \to \tilde{B}$, where $\tilde{B} \supset B$ is a bouquet of circles, such that its automorphism group remains unaltered, that is, $\mathrm{Aut}(\tilde{\Gamma}) \cong G$.
I am unsure how to build such an extension. Is there an easier way to proceed? Is the extension idea any good? Any help or hint would be highly appreciated.
I guess in order for this to be possible, you need a covering which is not regular. Let $\tilde B$ contain one more circle, and let $\tilde{\Gamma}$ be constructed as follows. Take a copy of $\Gamma$ and at every vertex attach a copy of $\mathbb R$ along $0\in \mathbb R$. To each of these new copies of $\mathbb R$, attach a copy of $B$ to each nonzero integer $n\in\mathbb R$. Now map the intervals $[n,n+1]$ in the copies of $\tilde{\Gamma}$ to the extra circle in $\tilde B$, map the copies of $B$ in the obvious way, and map $\Gamma$ by $p$. This is a non-regular covering space, as you can check. Translation along the extra copies of $\mathbb R$ has been killed, which means the automorphisms have to take place inside of $\Gamma$ alone.
Edit: A little simpler is to attach an extra circle to $\Gamma$ at all vertices except the basepoint. Attach a copy of $\mathbb R$ at the basepoint to $0\in\mathbb R$, and attach a copy of $B$ to the nonzero integers in $\mathbb R$. The extra circles map onto the extra circle in $\tilde B$ and everything else maps as above.