I have a given group, defined by its table, namely:
Now I am asked to find a set $X$ where this group acts upon non trivially. I have problems understanding this question.
As I understand, I am looking for a certain permutation representation on which this group acts. Now as far as group actions go, they are defined as $$\varphi: G \rightarrow S_{A},$$ where $S_{A}$ is the permutation of some set $A$ by some action (which is an element of $G$).
My initial plan was to take $B \in G$ and see where this maps all the other elements of $G$. Then I find $$\varphi(B): B \rightarrow A, C \rightarrow D, D \rightarrow C, E \rightarrow F, F \rightarrow E, G \rightarrow H, H \rightarrow G, I \rightarrow J, J \rightarrow I K \rightarrow L, L \rightarrow K, M \rightarrow N, N \rightarrow M, O \rightarrow P, P \rightarrow O.$$
Does it make sense to just replace the letters by numbers in the alphabet to create my permutation set? Thus I would define $X$ as
$$X=\{(12)(34)(56)(78)(9 10)(11 12)(13 14)(15 16) \},$$
or am I completely off track? Any suggestions, hints or insight would be much appreciated!
A quick remark before we begin: This seems like an exercise which is meant to reinforce the statement and proof of "Cayley's Theorem" (which says that you can realised every group as a group of permutations). So you should look up this theorem and its proof, and try to understand how it connects to both your question here and to my answer.
Lets begin: A group $G$ is a set $S$ with an operation $\cdot$, and we often write $G=(X, \cdot)$. As such, a group always acts on its underlying set via left multiplication, so the action of $g\in G$ on $G$ is defined as: $$x\mapsto g\cdot x.$$ We often shorten "$G$ acts on its underlying set" to "$G$ acts on itself". Note that $G$ also acts on itself via right multiplication, so $x\mapsto x\cdot g$ which is completely analogous to left multiplication, and via conjugation, so $x\mapsto g^{-1}xg$ which is completely different.
So to answer your questions:
Your group acts $G$ on the set $X=\{A, B, \ldots, P\}$ via left multiplication, as the set $X$ is the set underlying your group $G$. As $G$ is not the trivial group, the action is non-trivial. Stop here and you will get full marks*!
Yes, this makes sense, but is not necessary to answer the question. However, if you want to understand your group $G$ as a permutation on some numbers (and in the comments you seem to want to do this) then this idea is the way to go. So, as you did in your question, replace $X$ with $Y=\{1, \ldots, 16\}$ using the bijection $$1\leftrightarrow A, 2\leftrightarrow B, \ldots, 16\leftrightarrow P.$$ You can then understand the action of $G$ on $Y$ by applying this bijection to the Cayley table which you gave in the question. So, for example, $B\cdot1=2$ and $B\cdot2=1$.
You are off track, but not completely. The permutation $(1, 2)(3, 4)(5, 6)(7, 8)(9, 10)(11, 12)(13, 14)(15, 16)$ corresponds to the action of $B$ on the set $Y$. That is, $B=(1, 2)(3, 4)(5, 6)(7, 8)(9, 10)(11, 12)(13, 14)(15, 16)$. But its almost like you have gone too far: you have already found the set which your group acts on, you don't need to find the action explicitly!
*Although possibly expand on why the action is non-trivial.