Question

Suppose $G_1, \ldots, G_k$ are unitary, Hermitian, and anti-commuting, as well as $F_1, \ldots, F_k$. If they 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 finite group. If $\pi, \sigma$ are equivalent, i.e., there exists $T \in GL_n(\mathbb{C})$ such that $T \pi(g) = \sigma(g) T$ for all $g \in G$, is $\pi$ and $\sigma$ unitarily equivalent, that is, there exists $T' \in U_n$ such that $T' \pi = \sigma T'$?

Answer 1

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.

Answer 2

Two similar unitary representations are always unitarily similar if they are irreducible. An isomorphism $V_1 \to V_2$ pulls back the unitary metric on $V_2$ to a unitary metric on $V_1$. To make this unitary, we need to change that metric into the original metric on $V_1$ through a $G$-equivariant map $V_1\to V_1$. But any two unitary metrics on a representation are related by a $G$-equivariant map, so this is always possible.