What is the process behind finding a Cayley permutation representation. For example, let's find the Cayley permutation representation of $\mathcal D_3$ in $S_6$.
$\mathcal D_3 = \left<r,s \mid r^3=s^2=1, rs=sr^{-1}\right>$.
Write,
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
1 & r & r^2 & s & rs & r^2s
\end{pmatrix}
By multiplying on the right, we find:
$\varphi_1=()$
$\varphi_r=(123)(456)$
$\varphi_{r^2}=(132)(465)$
$\varphi_s=(14)(26)(35)$
$\varphi_{rs}=(15)(24)(36)$
$\varphi_{r^2s}=(16)(25)(34)$

What is the formal process behind finding a Cayley permutation representation. Is this a group action from $\mathcal D_3$ on $\mathbb Z_6$? A group action wouldn't give you cycles like this.

 A: This all looks good to me. You were looking for an injective homomorphism $\phi : \mathcal{D}_3 \to \operatorname{Sym}(\mathcal{D}_3) \cong S_6$ by letting $\mathcal{D}_3$ act on itself; and that's exactly what you've found. 
In general, you've followed the process: 


*

*Set up a bijection, $L : \{1, 2, \ldots, \lvert G\rvert \} \to G$, labeling the elements of your group

*For each group element $g \in G$, construct the permutation $\varphi_g = L^{-1} \circ r_g \circ L,$ which is a permutation in $S_{\lvert G \rvert}$ (here, the map $r_g : G \times G \to G$ is the right multiplication by $g$ map; I don't know if you used right- or left-multiplication, but it would  only be a superficial difference leading to isomorphic representations)

*Your injective homomorphism is then $\phi : G \to S_{\lvert G \lvert}$, given by $\phi(g) = \varphi_g$


I'm not sure I understand your comments about cycles, or $\Bbb Z_6$. You're just writing permutations in $S_6$ in cycle notation; of course there will be cycles. It doesn't mean that your group is cyclic (but again, I'm really not sure what that comment means).
As far as "a group action on $\Bbb Z_6$", I guess you could call it that, but it would be misleading. You could consider this a group action on any set of $6$ elements (by setting up a bijection like the $L$ used here), but traditionally you'll pick the integers from $1$ to the order of your group (or just the original group elements themselves). However, just because your group acts on a set in bijection with $\Bbb Z_6$, that doesn't mean this permutation representation ties them together in any way. It's strictly about cardinality here; groups act on sets, you need a homomorphism to $\Bbb Z_6$ (which this isn't) if you want to relate their algebraic properties.
