Orbit/stabiliser of an action Question. Suppose $X=\{\heartsuit,\diamondsuit,\clubsuit,\spadesuit\}$ and $\pi$ is an action of $\mathcal{D}_8=\{1,r,r^2,r^3,s,rs,r^2s,r^3s\}$ satisfying $$\pi_r(\heartsuit)=\spadesuit,\quad\pi_r(\diamondsuit)=\clubsuit,\quad\pi_r(\spadesuit)=\heartsuit,\quad\pi_r(\clubsuit)=\diamondsuit,$$ $$\pi_s(\heartsuit)=\diamondsuit,\quad\pi_s(\diamondsuit)=\heartsuit,\quad\pi_s(\spadesuit)=\spadesuit,\quad\pi_s(\clubsuit)=\clubsuit.$$
Find the orbit and stabiliser of $\diamondsuit$.

My initial thoughts was constructing $\pi_g$ for all $g\in\mathcal{D}_8$, for instance $$\pi_{rs}(\heartsuit)=\pi_r\circ\pi_s(\heartsuit)=\pi_r(\diamondsuit)=\clubsuit,$$ $$\pi_{rs}(\spadesuit)=\pi_r\circ\pi_s(\spadesuit)=\pi_r(\spadesuit)=\heartsuit,$$ $$\pi_{rs}(\diamondsuit)=\pi_r\circ\pi_s(\diamondsuit)=\pi_r(\heartsuit)=\spadesuit,$$ $$\pi_{rs}(\clubsuit)=\diamondsuit.$$
Incidentally, continuing this gives $$\pi_{r^2s}(\diamondsuit)=\heartsuit,\quad\pi_{r^2s}(\heartsuit)=\diamondsuit,\quad \pi_{r^2s}(\clubsuit)=\clubsuit,\quad\pi_{r^2s}(\spadesuit)=\spadesuit.$$ $$\pi_{r^2}(\diamondsuit)=\diamondsuit,\quad\pi_{r^2}(\heartsuit)=\heartsuit,\quad\pi_{r^2}(\spadesuit)=\spadesuit,\quad\pi_{r^2}(\clubsuit)=\clubsuit.$$ Should I find it strange that this coincides with $\pi_1(g)?$ $$\pi_{r^3}(\diamondsuit)=\clubsuit,\quad\pi_{r^3}(\heartsuit)=\spadesuit,\quad\pi_{r^3}(\spadesuit)=\heartsuit,\quad\pi_{r^3}(\clubsuit)=\diamondsuit.$$
$$\pi_{r^3s}(\diamondsuit)=\spadesuit,\quad\pi_{r^3s}(\spadesuit)=\heartsuit,\quad\pi_{r^3s}(\heartsuit)=\clubsuit,\quad\pi_{r^3s}(\clubsuit)=\diamondsuit.$$ $$\pi_1(\diamondsuit)=\diamondsuit,\quad\pi_1(\clubsuit),\quad\pi_1(\heartsuit)=\heartsuit,\quad\pi_1(\spadesuit)=\spadesuit.$$
So now, by definition, we have that the stabiliser of $x$ is the set of elements of $G$ that keep $x$ fixed under the action. From the above computations; we see $$\text{Stab}(\diamondsuit)=\{1, r^2\}.$$
Similarly, by definition, we have that the orbit of $x$ is the set of possible destinations $x$ goes to under the action. From the above computations; we see $$\text{Orb}(\diamondsuit)=\{\heartsuit, \spadesuit, \diamondsuit, \clubsuit\}.$$

I think my final answer is right; the Orbit-Stabiliser theorem seems to hold here.
So, while I guess the way I did it wasn't too exhausting an approach; but I feel like my approach is quite naive and that there's a more savvy way to do this. 
I'd be very grateful for a pointer in the right direction for me to be better at this type of question in the future.
Thanks in advance.
 A: One way of thinking of this is to think of the action as having elements of $D_8$ act on $X$ as permutations and just do your algebra there.
Basically, you can see from the definition of the action that $r$ acts as the permutation $(\heartsuit \spadesuit)(\diamondsuit \clubsuit)$ and $s$ acts as $(\diamondsuit \heartsuit)$. You can use what you know about Symmetric groups to compute the stabilizer. For instance, because $r$ acts as a products of $2$-cycles, it's clear that $r^2$ acts trivially. It's also clear that $r^3$ can't act trivially.
You can figure out the orbit of $\diamondsuit$ once you have everything in cycle notation by just applying combinations of $(\heartsuit \spadesuit)(\diamondsuit \clubsuit)$ and $(\diamondsuit \heartsuit)$ to $\diamondsuit$. I'm not sure how much faster this will be than what you did, but it's another way. Using Orbit Stabilizer Theorem to figure out how many things you should expect in the orbit was the right thing to do.
I think the biggest thing you should notice is that everything in the action is determined by the action of $r$ and $s$, since the group is generated by them, so usually you don't actually have to compute everything to figure out what's going on, you just have to understand really well what's happening with $r$ and $s$.
