Inner automorphisms of Pauli strings in the unitary subgroup of matrices Statement
What is the set $\mathcal{T}_n$ of matrices $T \in GL(2^n)$ such that for all Pauli strings $P \in \mathcal{P}_n=\{\otimes_{i=1}^n \sigma_{m_i}\mid m\in \{1,2,3\}^n, \sigma_{m_i} \text{ Pauli matrix}\} $, $T P T^{-1}$ is a unitary matrix, i.e. $T\mathcal{P}_nT^{-1} \subset U(2^n)$?
Possible direction
We know $U(2^n) \subset \mathcal{T}_n$, since all Pauli strings are unitary operators as well.
More generally, applying the QR decomposition on $T$, we can write $T = Q R$, for $Q \in U(2^n)$ and $R$ an upper triangular matrix. Hence, $T \in \mathcal{T}_n \Leftrightarrow R P R^{-1} \in U(2^n)$
Case $n=1$
For $n=1$, everything is a 2 by 2 matrix, so we can write $R = \pmatrix{a & c \\ 0 & b}$, $a,b \neq 0$. The conjugation of $\sigma_3 = \pmatrix{1 & 0 \\ 0 & -1}$ by $R$ is, then, $R \sigma_3 R^{-1} = \pmatrix{1 & -2 c / b \\ 0 & -1}$. The unitarity condition $R \sigma_3 R^{-1} \in U(2)$ implies $c = 0$. Doing the same for $\sigma_1$ and $\sigma_2$ reveals that $|a| = |b|$, so $\lambda R \in U(2)$ for $\lambda = \frac{1}{\sqrt{ab}} \in \mathbb{C}^*$. Hence, $\mathcal{T}_1 = \mathbb{C}^* \cdot U(2)$.
Conjecture
I conjecture that $\mathcal{T}_n = \mathbb{C}^* \cdot U(2^n)$. I checked that this is true for $n=2$ following the same line of argument of the $n=1$ case explained above.
 A: The answer is much simpler.
Firstly, we can expand $\mathcal{P}_n$ to the subgroup $\langle \mathcal{P}_n \rangle$ generated by it, which also includes terms with the identity matrix $1_{2\times2} = \sigma_i^2$.
Then, $T P T^{-1} \in U(2^n)$ for $P \in \langle \mathcal{P}_n \rangle$ implies
$$
\begin{align}
(TPT^{-1})^\dagger & = (T P T^{-1})^{-1} \\
(T^\dagger)^{-1} P T^{\dagger} & = T P T^{-1} \\
[P,T^\dagger T] & = 0
\end{align}
$$
where in the second line we used that $P$ is hermitian and unitary.
Since $\mathrm{span}_\mathbb{C}(\langle \mathcal{P}_n \rangle) = M_\mathbb{C}(2^n)$, the vector space of all matrices, then $T^\dagger T \in Z(M_\mathbb{C}(2^n)) = \mathbb{C} \cdot 1_{2^n \times 2^n}$.
Using the polar decomposition of $T = U R$, where $U$ is unitary and $R$ is positive semidefinite, the last conclusion implies $R \in \mathbb{R}^+ \cdot 1_{2^n \times 2^n}$ and thus $T \in \mathbb{R}^+ \cdot U(2^n)$.
EDIT: We don't even have to assume that $T$ is invertable. If for every $P \in \mathbb{P}_n$, $T P = U_P T$ for some $U_P$, then taking the hermitian conjugate gives $P T^\dagger = T^\dagger U_P$, so
$$
P T^\dagger T = T^\dagger U_P T = T^\dagger T P.
$$
