$\DeclareMathOperator\GL{GL}$Let $G$ be a subgroup of $\GL_n(\mathbb{C})$ such that for every $g \in G$ there exists $c \in \GL_n(\mathbb{C})$ for which $cgc^{-1}$ is unitary (or, which is the same, $g$ is diagonalizable with unimodular eigenvalues). Does there exist $c \in \GL_n(\mathbb{C})$ such that $cGc^{-1}$ consists of unitary matrices?

A similar question is discussed in Venkataramana's answer to Groups of matrices in which all elements have all eigenvalues equal in modulus, and this way I can obtain that $G$ is isomorphic to a subgroup of the unitary group.

Let $\mathbb{R}^n = V$, and $0 = V_0 \subset V_1 \subset V_2 \subset \dotsb \subset V_k = V$ be subspaces such that each $V_i$ is invariant under $G$ and the induced action of $G$ on $V_{i+1}/V_i$ is irreducible. Then I get a homomorphism: $G \to \GL(V_1 \oplus (V_2/V_1) \oplus \dotsb \oplus (V_{k}/V_{k-1}))$, and (see the link above) I can assume that its image consists of unitary matrices. Its kernel is trivial, since this homomorphism preserves eigenvalues and the kernel consists of diagonalizable matrices with all eigenvalues equal to $1$.

Now I see two ways:

This homomorphism (denoted as $S$ and seen as a bijection: $G \to \operatorname{Im}S$) is continuous. If $S^{-1}$ is continuous, I think I'm able to prove that $G$ is bounded. But I'm not sure that it's continuous.

$\operatorname{tr}(ab)$ was a non-degenerate form on $\operatorname{Mat}_{n \times n}(\mathbb{C})$. Will it be non-degenerate on a subspace generated by $G$?

Something else….