Suppose $G\_1, \ldots, G\_k$ are unitary, Hermitian, and anti-commuting matrices, and assume the same for $F\_1, \ldots, F\_k$. Suppose these matrices are similar, i.e. there exists $T \in GL\_n(\mathbb{C})$ such that
$$
G\_i = T^{-1} F\_i T
$$
for all $i \in [k]$. Does there exist $V \in U\_n$, where $U\_n$ is the group of unitary matrices, such that
$$
G\_i = V^{-1} F\_i V
$$
for all $i \in [k]$?


**Added later:**  

Generally, let $\pi, \sigma : G \to U\_n(\mathbb{C})$ be two unitary representations of a finite group. Suppose $\pi$, $\sigma$ are equivalent, i.e. there exists $T \in GL\_n(\mathbb{C})$ such that $T \pi(g) = \sigma(g) T$ for all $g \in G$. Are $\pi$ and $\sigma$ unitarily equivalent, that is, there exists $T' \in U\_n$ such that $T' \pi = \sigma T'$?



Similarity about unitary matrices
Maybe this question deserves an answer which doesn't use decomposition into irreducible representations. Let $\pi$ and $\sigma$ be equivalent unitary representations. If $T$ intertwines, then so does $T^{\ast}$, because $\pi(g^{-1})=\pi(g)^{\ast}$ (same for $\sigma$). Now we can conclude that $|T|=\sqrt{T^{\ast}T}$ is also an intertwiner. Finally, $T|T|^{-1}$ is the unitary we need.


