Let us consider the vector space of complex $n \times n$ matrices.
Let $\{ V_i \}_{i=1,2,\cdots,n^2}$ be a trace-orthogonal basis of matrices, i.e., $$ \forall i,j \in \{1,2,\cdots,n^2\} : \quad \textrm{tr}(V_i^\dagger V_j) = n \; \delta_{ij}. $$
In general, such matrices do not need to be unitary (e.g., the Gell-Mann matrices are a nice example for $n=3$). However, let us presume an additional condition: there exists an invertible matrix $M$ such that $$ \forall i \in \{1,2,\cdots,n^2\}: \quad M V_i M^{-1} \textrm{is a unitary matrix}. $$ In particular, this implies that each $V_i$ (and in fact arbitrary products $V_{i_1} V_{i_2} \cdots V_{i_k}$) are similar to unitary matrices.
Question: Do the above two properties imply that each $V_i$ is unitary?
Assuming the dagger symbol means conjugate transpose, then the answer is yes.
The fact that $M V_i M^{-1}$ is unitary means arbitrary $V_i$ has all eigenvalues on the unit circle. And
$n=\textrm{trace}\big(V_i^\dagger V_i\big) =\big \Vert V_i\big\Vert_F^2 \geq \sum_{k=1}^n \vert \lambda_k^{(i)}\vert^2=\sum_{k=1}^n 1 =n$
$\implies$ each $V_i$ is normal since Schur's Inequality is met with equality. Applying spectral theorem we have $V_i =Q_i D_i Q_i^{-1}=Q_i D_i Q_i^\dagger$ i.e. $V_i$ is the product of 3 unitary matrices hence it is unitary.