$\newcommand\dag\dagger$
I would assume that the property of being unitary is invariant under similarity transformations since similarity transformations are just a change of basis of a linear map, and if the matrix is unitary the map must be, so that every matrix that represents the map in some basis is unitary but I can't seem to prove it. Here's my attempt so far:
If $D$ is similar to a unitary matrix $U$ through a similarity matrix $S$ then $SDS^{-1}=U$ is unitary so $D^\dag D = (S^{-1}US)^\dag (S^{-1}US)$. But the only way I can think of making this expression equal to the identity matrix is if $S$ is unitary. Is my hypothesis correct? Can my first attempt be finished to prove my hypothesis?
Thanks
Suppose your n x n $D$ is diagonal with all diagonal values distinct and on the unit circle. $D$ is thus unitary.
Now suppose $S$ to be upper triangular with all ones on the diagonal and all ones above the diagonal. If your statement were true then you'd have unitary $U=SDS^{-1}$ but $U$ does not have eigenvectors that may be chosen to be mutually orthonormal. This contradicts spectral theorem since all unitary matrices are normal.
addendum
for a very simple example of this, consider
$A = \begin{bmatrix} 0 & k\\ \frac{1}{k} & 0 \end{bmatrix}$
with positive number $k$
$\text{trace}(A) = 0$
$\text{det}(A) = -1$
so it is similar to a diagonal matrix
$D = \begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}$
$D$ is unitary (orthogonal in reals)
but
$\big \Vert A \big \Vert_F^2 = k^2 + \frac{1}{k^2} \geq 2$
with equality iff k= 1, otherwise it is impossible for $A$ to be unitary.