Confusion about Actions of the Symmetric Group I’m working on some practice questions and I am having trouble understanding actions of the symmetric group.
I have the answers, but there were no explanations as to how they were derived. I feel that there is something very fundamental that i am not understanding.

$(i)$ Let $X=\{ \{i,j,k\} \subset \{1,2,3,4\} \mid |\{i,j,k\}| = 3 \} = $ the set of all $3$ element subsets. What is the number of orbits of $\mathrm{Sym}(4) \curvearrowright X$?

The answer here was given as $1$.
However, I don’t understand this. I have that the elements of $X$ are $\{123\},\{124\},\{234\},\{134\}$ (i.e., choosing combinations of $3$ from $4$).
For example, let $g_1 = (12)(34) \in \mathrm{Sym}(4)$ and say $x_1 = (123) \in X$.
Then $$g_1 \cdot x_1 = (214) \in X \,.$$ (Is this correct? $g_1$ sends $1 \rightarrow 2$, $2 \rightarrow 1$, $3 \rightarrow 4$?)
Similarly let $g_2 = (1234)  \in  \mathrm{Sym}(4)$, then
$$  g_2 \cdot x_1 = (234) \in X \,.$$
So from these two examples, already the orbit of $x_1$ under $G$ are two other elements in $X$, so how can the number of orbits be $1$?

$(ii)$ Note that Sym($n$) acts on the set of all the subsets of $\{1,\dotsc,n\}$ denoted $\rho(\{1,\dotsc,n\})$.  Let $X = \rho( \{1,\dotsc,4\})$. What is the number of orbits of
$\mathrm{Sym}(4) \curvearrowright X$?

The answer provided here is $5$. Again, I have a misunderstanding here, which is similar to above. Any insight would be greatly appreciated!
 A: You are talking about size of orbit, not the number of orbits. $X$ is the whole orbit which has $4$ elements.
For the second part, each orbit of $\rho(\{1,\dots,4\}$ contains subsets of $\{1,\dots,4\}$ of the same size. The number of orbits is equal to the number of possible sizes of the subsets which are $0,1,2,3,$ and $4$.
A: I don't think an answer can be given without specifying what action we are talking about. As for $(i)$, let:
$$\sigma*\{i,j,k\}:=\{\sigma(i),\sigma(j),\sigma(k)\}$$
You can prove that this is indeed an action of $S_n$ on $X$, for any $n\ge 3$. Next:
\begin{alignat}{1}
\operatorname{Stab}_*(\{i,j,k\}) &= \{\tau\in S_4\mid \tau\cdot\{i,j,k\}=\{i,j,k\}\} \\
&= \{\tau\in S_4\mid \{\tau(i),\tau(j),\tau(k)\}=\{i,j,k\}\} \\
&= \{\tau\in S_4\mid \tau(l)=l, \text{for }l\in \{1,2,3,4\}\setminus\{i,j,k\}\} \\
&= \operatorname{Stab}(l)
\end{alignat}
where the latter stabilizer is the one by $l\space(\ne i,j,k)$ of the natural action of $S_4$ on $\{1,2,3,4\}$. But this has clearly order $(4-1)!=6$, whence the $*$-orbit by $\{i,j,k\}$ has size $\frac{4!}{6}=4=|X|$, and hence there is one single $*$-orbit.
This argument can be generalized this way: each cardinality of the subsets of $S:=\{1,2,3,4\}$ corresponds to one orbit of the $*$-action of $S_4$ on the power set $\mathcal{P}(S)$. So, $5$ orbits overall, corresponding to the sizes $0$, $1$, $2$, $3$, $4$.
A: The first action is transitive, so that the number of orbits is one.
The second action is on $X=\mathscr P(\{1,2,3,4\})$, the set of subsets (power set) of $\{1,2,3,4\}$.  But, $S_4$ acts via bijections.  So it takes sets of given size to sets of the same size. It also transitive on each set of subsets of given size.
Thus the number of orbits is the number of possible sizes of subsets.   For a set of order $4$, that's $5$.
More generally,  if $S_n$ acts on $\mathscr P(\{1,2,\dots, n\})$, in the same way, the number of orbits is $n+1$.
