[example: Rotations of a Square and a Chosen Vertex]
Let $G$ be the rotational symmetry group of a square, so
\begin{align*}
G=C_4=\{e,r,r^2,r^3\},
\end{align*}
where $r$ is rotation by $90$ degrees counterclockwise and $r^4=e$. Label the vertices cyclically by $x_0,x_1,x_2,x_3$, and choose $x=x_0$. Then
\begin{align*}
e\cdot x_0 &= x_0, \\
r\cdot x_0 &= x_1, \\
r^2\cdot x_0 &= x_2, \\
r^3\cdot x_0 &= x_3.
\end{align*}
Hence the orbit is
\begin{align*}
G\cdot x_0
&= \{g\cdot x_0 : g\in G\} \\
&= \{e\cdot x_0,r\cdot x_0,r^2\cdot x_0,r^3\cdot x_0\} \\
&= \{x_0,x_1,x_2,x_3\}
= X.
\end{align*}
For the stabiliser, we test the same four elements:
\begin{align*}
e\cdot x_0 &= x_0, \\
r\cdot x_0 &= x_1 \ne x_0, \\
r^2\cdot x_0 &= x_2 \ne x_0, \\
r^3\cdot x_0 &= x_3 \ne x_0.
\end{align*}
Therefore
\begin{align*}
G_{x_0}
&= \{g\in G : g\cdot x_0=x_0\} \\
&= \{e\}.
\end{align*}
Now let $G$ act on the set $Y$ of the two diagonals. Write
\begin{align*}
d_1 &= \{x_0,x_2\}, &
d_2 &= \{x_1,x_3\}.
\end{align*}
The action on $d_1$ is
\begin{align*}
e\cdot d_1 &= \{e\cdot x_0,e\cdot x_2\}=\{x_0,x_2\}=d_1, \\
r\cdot d_1 &= \{r\cdot x_0,r\cdot x_2\}=\{x_1,x_3\}=d_2, \\
r^2\cdot d_1 &= \{r^2\cdot x_0,r^2\cdot x_2\}=\{x_2,x_0\}=d_1, \\
r^3\cdot d_1 &= \{r^3\cdot x_0,r^3\cdot x_2\}=\{x_3,x_1\}=d_2.
\end{align*}
Thus
\begin{align*}
G_{d_1}
&= \{g\in G : g\cdot d_1=d_1\} \\
&= \{e,r^2\}.
\end{align*}
The chosen vertex has no residual rotational symmetry, while the chosen diagonal is still preserved by the half-turn $r^2$.
[/example]