Complex-Orthogonal Similarity First, a quick definition: I say that a matrix $M$ with complex entries is complex-orthogonal if $MM^T = I$.  Note that $T$ here refers to the transpose, not the conjugate transpose.  That is, we are taking the adjoint with respect to a bilinear (as opposed to sesquilinear) dot-product over $\Bbb C$.  A complex-orthogonal matrix need not be unitary, and vice versa (unless of its entries are in $\Bbb R$).
Consider the following claim:

Claim: For any $A \in \Bbb C^{n \times n}$, there exists a complex-orthogonal $U$ such that $UAU^T$ has equal diagonal entries.

My Question: Does this claim hold?
It may be helpful to take a look at this paper which proves that for any $A$, there is an $S$ such that $SAS^{-1}$ has equal diagonal entries.  It goes on to show that such an $S$ may be taken to be a unitary matrix.  Another proof of the same fact is presented in p. 77 of R.A. Horn and C.R. Johnson’s Matrix Analysis (1985/7, Cambridge Univ. Press).  
Notably, both of these proofs exploit facts for which an analog fails to exist in my case.  For Horn and Johnson's proof, we need the fact that the unitary matrices form a compact subset of $\Bbb C^{n \times n}$.  For the proof linked, we need the numerical range to have certain "nice properties", which I believe fail in my case.
Any insight here is appreciated.
 A: The claim is false. As a counterexample to the claim consider
$$
A = \begin{pmatrix}
i & 1
\\
1 & -i
\end{pmatrix}.
$$
To see this, note that the $2 \times 2$ complex orthogonal matrices can be parameterised as
$$
O = 
\begin{pmatrix}
\cos z & \mp \sin z
\\
\sin z & \pm \cos z
\end{pmatrix}
$$
for $z \in \mathbb{C}$. This yields
$$
B = O A O^T = \mathrm{e}^{ \pm 2 i z}\begin{pmatrix}
i & \pm1
\\
\pm1 & -i
\end{pmatrix}.
$$
thus we always have $B_{11} = - B_{22}$, the only possible solution to $B_{11} = B_{22}$ is $B_{11} = 0 = B_{22}$, which is approached only in the limit $\mathrm{Im}(z) \to \pm \infty$, a limit in which every element of $O$ tends to infinity and $O$ is no longer invertable.
A: An easier justification of ComptonScattering's counterexample: since $A=\pmatrix{i&1\\ 1&-i}$ is nilpotent and complex symmetric, if $UAU^T$ has equal diagonal entries, we must have
$$
UAU^T=a\pmatrix{0&1\\ 1&0}
$$
for some $a\in\mathbb C$. But then $0=U(A^2)U^T=(UAU^T)^2=a^2I$. In turn, we get $a=0$ and $A=0$, which is a contradiction.
