Subgroups and invariants in a unitary group $U(2)$ mixing multiplets This is related to the post, but a simplified version of the problem. However, the form of $P_x,P_y,P_z$ are different and richer.
Let $$G=U(2),$$ be the unitary group. Here we consider $G$ in terms of the fundamental representation of U(2). Namely, all of $g \in G$ can be written as a rank-2 (2 by 2) matrices.
Can we find some subgroup of Lie group,  $$k \in K \subset G= U(2) $$  such that

$$ 
k^T \{P_x, P_y,  P_z, -P_x, - P_y, - P_z\} k =\{P_x, P_y,  P_z, -P_x, - P_y, - P_z\}.
$$
  This means that set $\{P_x, P_y,  P_z, -P_x, - P_y, - P_z\}$ is invariant under the transformation by $k$. 
  Here $k^T$ is the transpose of $k$.
  What is the full subset (or subgroup) of $K$?

Here we define: 
$$
P_x =
\left(
\begin{array}{ccc}
 0 & 1 \\
1 & 0  \\
\end{array}
\right),\;\;\;\; P_y =
\left(
\begin{array}{ccc}
 -1 & 0  \\
 0 & 1  \\
\end{array}
\right),\;\;\;\; P_z =
\left(
\begin{array}{ccc}
 -i & 0  \\
 0 & -i  \\
\end{array}
\right).$$
This means that $k^T P_a k= \pm P_b$ which may transform $a$ to a different value $b$, where $a,b \in \{x,y,z \}$. But overall the full set $ \{P_x, P_y,  P_z, -P_x, - P_y, - P_z\}$ is invariant under the transformation by $k$.
There must be a trivial element $k=$ the rank-2 identity matrix. But what else can it allow? 

How could we determine the complete $K$?

 A: 
Answer. The subgroup $K$ of invariant matrices in $U(2)$ is isomorphic to the finite group
  $$
\operatorname{SL}(2,3)\to G\to\mathbb{Z}_2) \rtimes_\varphi \mathbb{Z}_2
\quad\simeq\quad
\mathbb{Z}_2\to G\to S_4) \rtimes_\varphi \mathbb{Z}_2
$$
  where $\operatorname{SL}(2,3)$ is the special linear group of degree 2 over a field of 3 elements (order 24), which is isomorphic to the binary von Dyck group with the parameters (2,3,3); it has the presentation: $\langle a,b,c \mid a^3=b^3=c^2=abc\rangle$.
In other words, $K\cong G\rtimes_\varphi\mathbb{Z}_2$ for some subgroup $G\subset K$ and $G/\operatorname{SL}(2,3)\cong\mathbb{Z}_2$ or $G/\mathbb{Z}_2\simeq S_4$.

For whom is interested, I list the all $96=24\times 4$ invariant matricies of $K$:
$$
\begin{gather*}
\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},
\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},
\frac{(1+i)}{\sqrt{2}} \begin{pmatrix} 1 & 0 \\ 0 & -i \end{pmatrix},
\frac{(1-i)}{\sqrt{2}} \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix}, \\
\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},
\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix},
\frac{(1+i)}{\sqrt{2}} \begin{pmatrix} 0 & 1 \\ -i & 0 \end{pmatrix},
\frac{(1-i)}{\sqrt{2}} \begin{pmatrix} 0 & 1 \\ i & 0 \end{pmatrix}, \\
\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix},
\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix},
\frac{(1+i)}{2} \begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix},
\frac{(1-i)}{2} \begin{pmatrix} 1 & 1 \\ -i & i \end{pmatrix}, \\
\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ -1 & -1 \end{pmatrix},
\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix},
\frac{(1+i)}{2} \begin{pmatrix} 1 & -1 \\ i & i \end{pmatrix},
\frac{(1-i)}{2} \begin{pmatrix} 1 & -1 \\ -i & -i \end{pmatrix}, \\
\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -i \\ -i & 1 \end{pmatrix},
\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -i \\ i & -1 \end{pmatrix},
\frac{(1+i)}{2} \begin{pmatrix} 1 & -i \\ -1 & -i \end{pmatrix},
\frac{(1-i)}{2} \begin{pmatrix} 1 & -i \\ 1 & i \end{pmatrix}, \\
\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & i \\ -i & -1 \end{pmatrix},
\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix},
\frac{(1+i)}{2} \begin{pmatrix} 1 & i \\ -1 & i \end{pmatrix},
\frac{(1-i)}{2} \begin{pmatrix} 1 & i \\ 1 & -i \end{pmatrix}
\end{gather*}
$$
product with $\langle i\rangle=\{\pm I,\pm iI\}$.
For given orders, the numbers of elements in $K$ and $\mathbb{Z}_4\times S_4$ are as follows:
order | K        | Z4 x S4
--------------------------
1     | 1        | 1
2     | 19       | 19
3     | 8        | 8
4     | 20       | 44
6     | 8        | 8
8     | 24       | 0
12    | 16       | 16
--------------------------
total | 96       | 96


This information shows that $K$ (GAP ID [96, 192]) in $U(2)$ is not isomorphic to the previous answer $\mathbb{Z}_4\times S_4$ (GAP ID [96,186]) in $U(3)$.

