Invariant subgroup in a Hermitian self-conjugate pair multiplication in Unitary Group $U(3)$ Inspired by @ChoF's exploration, I like to post a similar but possibly richer structure related to the decompositions of $U(3)$ or $SU(3)$ representation multiplications $3 \otimes \bar 3 = 1 \oplus  8.$
3 is in the fundamental and 8 is in the adjoint representation of $SU(3)$. Below I modify and preserve the structure of questions from (Invariant subgroup in a self-conjugate pair multiplication in Unitary Group $U(3)$)
Let $$G=U(3),$$ be the unitary group. Here we consider $G$ in terms of the fundamental representation of U(3). Namely, all of $g \in G$ can be written as a rank-3 (3 by 3) matrices. Now we take a set of $P$ matrices from Gell-Mann matrices and generalized Pauli matrices. And all  $P$ matrices are Hermitian.
Can we find some subgroup of Lie group,  $$k \in K \subset G= U(3) $$  such that

$$ 
k^\dagger \{\pm P_1, \pm  P_2,\pm   P_3,\pm  P_4,\pm  P_5, \pm  P_6, \pm  P_7, \pm  P_8, \pm  P_9 \} k$$ 
  $$=\{\pm P_1, \pm  P_2,\pm   P_3,\pm  P_4,\pm  P_5, \pm  P_6, \pm  P_7, \pm  P_8, \pm  P_9\}.
$$
  This means that the full set $\{\pm P_1, \pm  P_2,\pm   P_3,\pm  P_4,\pm  P_5, \pm  P_6, \pm  P_7, \pm  P_8, \pm  P_9\}$ is invariant under the transformation by $k$. 
  Here $$k^\dagger \equiv (k^*)^T$$ is the complex conjugate ($*$) transpose ($T$) of $k$.
  What is the full subset (or subgroup) of $K \subset G$?

Here we define: 
$$
P_1 =
\left(
\begin{array}{ccc}
 0 & 1 & 0 \\
1 & 0 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right),\;\;\;\; P_2 =
\left(
\begin{array}{ccc}
 0 & -i & 0 \\
 i & 0 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right),\;\;\;\; P_3 =
\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & -1 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right)
$$
$$
P_4 =
\left(
\begin{array}{ccc}
 0 & 0 & 1 \\
0 & 0 & 0 \\
 1 & 0 & 0 \\
\end{array}
\right),\;\;\;\; P_5 =
\left(
\begin{array}{ccc}
 0 & 0 & -i \\
0 & 0 & 0 \\
 i & 0 & 0 \\
\end{array}
\right),\;\;\;\;
P_6 =
\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
0 & 0 & 0 \\
 0 & 0 & -1 \\
\end{array}
\right)$$ 
$$P_7 =
\left(
\begin{array}{ccc}
 0 & 0 & 0 \\
0 & 0 & 1 \\
 0 & 1 & 0 \\
\end{array}
\right),\;\;\;\; P_8 =
\left(
\begin{array}{ccc}
 0 & 0 & 0 \\
0 & 0 & -i \\
 0 & i & 0 \\
\end{array}
\right),\;\;\;\;
P_9 =
\left(
\begin{array}{ccc}
 0 & 0 & 0 \\
0 & 1 & 0 \\
 0 & 0 & -1 \\
\end{array}
\right).$$
This means that $$k^\dagger P_a k= \pm P_b$$ which may transform $a$ to a different value $b$, where $a,b \in \{1,2,3,4,5,6,7,8,9 \}$. But overall the full set $ \{\pm P_1, \pm  P_2,\pm   P_3,\pm  P_4,\pm  P_5, \pm  P_6, \pm  P_7, \pm  P_8, \pm  P_9 \}$ is invariant under the transformation by $k$.

Please determine the complete $K$.

p.s. I modify the Gell-Mann matrices to a most symmetric form in accordance.
 A: The nine matrices $P_i$ can be divided into three sets:
$$
\mathcal P_1=\{\pm P_1,\pm P_2,\pm P_3\}, \quad
\mathcal P_2=\{\pm P_4,\pm P_5,\pm P_6\}, \quad
\mathcal P_3=\{\pm P_7,\pm P_8,\pm P_9\}.
$$
According to the transformation $k^\dagger\mathcal P_i k$, we have 6 cases as follows:


*

*(All fixed) If $\mathcal P_1\leftrightarrow\mathcal P_1$, $\mathcal P_2\leftrightarrow\mathcal P_2$, $\mathcal P_3\leftrightarrow\mathcal P_3$, then we have
$$
k = \begin{pmatrix} \alpha & 0 & 0 \\ 0 & \beta & 0 \\ 0 & 0 & \gamma \end{pmatrix} \in U(1)\times U(1)\times U(1)
$$

*($\mathcal P_1$ fixed) If $\mathcal P_1\leftrightarrow\mathcal P_1$, $\mathcal P_2\leftrightarrow\mathcal P_3$, then we have
$$
k = \begin{pmatrix} 0 & \alpha & 0 \\ \beta & 0 & 0 \\ 0 & 0 & \gamma \end{pmatrix} \in U(1)\times U(1)\times U(1)
$$

*($\mathcal P_2$ fixed) If $\mathcal P_2\leftrightarrow\mathcal P_2$, $\mathcal P_1\leftrightarrow\mathcal P_3$, then we have
$$
k = \begin{pmatrix} 0 & 0 & \alpha \\ 0 & \beta & 0 \\ \gamma & 0 & 0  \end{pmatrix} \in U(1)\times U(1)\times U(1)
$$

*($\mathcal P_3$ fixed) If $\mathcal P_3\leftrightarrow\mathcal P_3$, $\mathcal P_1\leftrightarrow\mathcal P_2$, then we have
$$
k = \begin{pmatrix} \alpha & 0 & 0 \\ 0 & 0 & \beta \\ 0 & \gamma & 0  \end{pmatrix} \in U(1)\times U(1)\times U(1)
$$

*(No fixed) If $\mathcal P_1\rightarrow\mathcal P_2\rightarrow\mathcal P_3\rightarrow\mathcal P_1$, then we have
$$
k = \begin{pmatrix} 0 & 0 & \alpha \\ \beta & 0 & 0 \\ 0 & \gamma & 0  \end{pmatrix} \in U(1)\times U(1)\times U(1)
$$

*(No fixed) If $\mathcal P_1\rightarrow\mathcal P_3\rightarrow\mathcal P_2\rightarrow\mathcal P_1$, then we have
$$
k = \begin{pmatrix} 0 & \alpha & 0 \\ 0 & 0 & \beta \\ \gamma & 0 & 0 \end{pmatrix} \in U(1)\times U(1)\times U(1)
$$
For all six cases, $\alpha,\beta,\gamma\in U(1)$ must satisfy
$$
(\alpha=\pm\gamma \text{ or} \pm i\gamma) \quad\text{and}\quad
(\beta=\pm\gamma \text{ or} \pm i\gamma)
$$

Answer. The invariant subgroup $K$ of $U(3)$ is isomorphic to the infinite group
  $$
(U(1) \times \mathbb{Z}_4\times\mathbb{Z}_4) \rtimes_\varphi S_3
$$

Note. The answer is quite related to the previous one.
