Constructing 6-fold cover of $S^1 \vee S^1$ with deck transformation group $\cong S_6$ So i'm thinking that this will be a cover space of maximum possible symmetry. Will a "necklace" of 6 circles work? Any tips appreciated

There is a $6!$-fold connected normal cover of $S^1\vee S^1$ with deck transformation group $S_6$.  Recall that $S_6$ has a presentation with two generators, hence there is a normal subgroup $N\subset F_2$ of the free group on two generators.  Let's identify $F_2=\pi_1(S^1\vee S^1)$.  Hence, there is a homomorphism $f:\pi_1(S^1\vee S^1)\to S_6$ through the quotient map $F_2\to F_2/N$.
By standard facts about covering spaces (as seen in Hatcher) there is then a connected normal cover of $S^1\vee S^1$ associated to the normal subgroup $N$ for which $S_6$ is the group of deck transformations.
You might know this covering space as the Cayley graph of $S_6$ with the chosen generating set.  In fact, the two-generator Cayley graphs of $S_6$ encompass the set of all connected normal covers of $S^1\vee S^1$ with deck transformation group isomorphic to $S_6$.
(William in the comments points out that there is a $6$-fold disconnected cover of $S^1\vee S^1$ with $S_6$ as its group of deck transformations: the disjoint union $\coprod_{i=1}^6S^1\vee S^1$.)
