Can the covering transformation group of a 6-sheeted covering map $p : X \to Y$ have exactly 12 elements?
I suspect that the answer is negative, but I cannot see an invariant that shows this. We don't have any limitations on the nature of spaces involved (i.e. they are not necessarily CW-complexes, etc.).
Edit: I almost completely rewritten my original answer.
Let $p: X\to Y$ be a covering map of topological spaces. The group of deck-transformations $Deck(p)$ is the subgroup of $Homeo(X)$ consisting of all homeomorphisms $g$ such that $p\circ g=p$. In particular, such a group has to act freely on $X$, provided that $X$ is connected. Now, pick a point $x\in X$. Then $p$ is constant on the $G$-orbit $G\cdot x$, which projects to a point $y\in Y$. This orbit has cardinality equal $|G|$. Therefore, $$ d=|p^{-1}(y)|\ge |G|$$ where $|Z|$ denotes the cardinality of a set $Z$. Thus, for connected $X$, one cannot have $|G|=12, d=6$.
For disconnected $X$ the picture is much messier. Here is an example with $|G|=12, d=6$:
Let $p_1: X_1\to Y_1$ be a degree 6 cyclic regular cover (say, circle covering itself). The deck-group of $p_1$ is $G_1=Z_6$. Let $p_2: X_2\to Y_2$ be a degree 6 covering map with the deck-group $G_2=Z_2$. To construct such, note that if $H$ is the dihedral group of order $12$ and $H'<H$ is the order 2 reflection subgroup, then the normalizer of $H'$ in $H$ is $Z_2\times Z_2$ and, hence, $H'$ is index 2 subgroup in its normalizer. Furthermore, $6=|H:H'|$. Now, consider a connected complex $Y_2$ with the fundamental group $H$ and let $p_2: X_2\to Y_2$ denote the 6-fold covering map associated with the subgroup $H'<H$ as above. Then its deck-transformation group is $Z_2$.
Lastly, let $X=X_1\sqcup X_2$, $Y=Y_1\sqcup Y_2$, $p: X\to Y$ be the map whose restriction to $X_i$ is $p_i$. Then $p$ is a 6-fold covering; its deck-transformation group is $G_1\times G_2$, it has order $12=6\times 2$.