$\DeclareMathOperator\GL{GL}$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](https://mathoverflow.net/a/244208/505475) to [Groups of matrices in which all elements have all eigenvalues equal in modulus](https://mathoverflow.net/questions/241565/groups-of-matrices-in-which-all-elements-have-all-eigenvalues-equal-in-modulus/244208#244208), 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](https://mathoverflow.net/a/244208/505475)) 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:


1. 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.
2. $\operatorname{tr}(ab)$ was a non-degenerate form on $\operatorname{Mat}\_{n \times n}(\mathbb{C})$. Will it be non-degenerate on a subspace generated by $G$?
3. Something else….



Group of matrices in which every matrix is similar to unitary
An example in $GL(3, {\mathbb C})$ was first given by Bass (answering a question by Kaplansky) in Example 1.10 of


*Bass, Hyman*, [**Groups of integral representation type**](https://doi.org/10.2140/pjm.1980.86.15), Pac. J. Math. 86, 15-51 (1980). [ZBL0444.20006](https://zbmath.org/?q=an:0444.20006).


In section 1 of his paper Bass also discusses the general structure of subgroups of $GL(n, {\mathbb C})$ where every element is unitarizable.


Here is the example. Start with a free subgroup $F=\langle s, t\rangle$ of rank 2 in $SU(2)$ acting linearly on ${\mathbb C}^2$ with generators acting as matrices $A, B$. Now, deform $F$ to a free group of affine transformations by
$$
\rho(s)=A, \rho(t)x= Bx+ v,
$$
where $v$ is any nonzero vector in ${\mathbb C}^2$. Then each element of $G=\rho(F\_2)$ is affine-conjugate to its linear part, an element of $SU(2)$. At the same time, $G$ has no fixed points in the complex-affine space ${\mathbb C}^2$. It follows that $G$ is unbounded. Lastly, use the fact that the group of complex-affine transformations of ${\mathbb C}^2$ embeds in $GL(3, {\mathbb C})$ sending an affine transformation
$$
x\mapsto Mx + v
$$
to the matrix
$$
\left[\begin{array}{cc}
M&v\\
0&1
\end{array}\right].
$$
Under this map unbounded subsets map to unbounded subsets.
This gives an example of a non-unitarizable subgroup of $SL(3, {\mathbb C})$ such that each individual element is unitarizable.


