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

The comments are correct that this construction is impossible, however the reason is flawed.
First of all, the action of the deck transformation group $G$ of a connected covering map $f : X \to Y$ a free action, meaning that the action of a nontrivial element of $G$ has no fixed points --- so no, a deck transformation cannot take $y_2$ to $y_3$ whilst fixing $y_1$. 
Second, given $x \in X$ and $y=f(x) \in Y$, the action of $G$ on the subset $f^{-1}(y) \subset X$ need not be transitive. In fact, that action is transitive if and only if the covering map $f : X \to Y$ is regular, if and only if the image of the induced homomorphism $\pi_1(X,x) \to \pi_1(Y,f(x))$ is a normal subgroup of $\pi_1(Y,f(x))$. However, this only makes the construction more impossible (if that means anything), because it implies that the cardinality of $f^{-1}(x)$ is greater than or equal to the order of the group $G$.
So, in a covering map of degree $6$, the set $f^{-1}(x)$ has cardinality $6$, and there is no free action of a group of order $6! = 120$ on a set of cardinality $6$.
For proofs of the various things discussed in this answer, I suggest sitting down with a good solid course on the relationship between fundamental groups and covering spaces, as in Hatcher.
