Condition for a Hermitian matrix to fulfill $H^\top = U \cdot H\cdot U$ for some symmetric unitary $U$. I am wondering about the following mathematical question, motivated by a physics problem from topological band theory (a bit more on that at the very end). For now let me strip all the unimportant physics jargon.
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Given a Hermitian matrix $H$ (i.e. $H = H^\dagger$), the question is whether there is a symmetric unitary matrix $U$ (i.e. $U = U^\top$ and $U\cdot U^\dagger = \mathbf{1}$) such that $H^\top = U\cdot H \cdot U^\dagger$ (Note: in my original question, the dagger at $U$ has been missing -- my mistake). The Hermitian matrix $H$ can be assumed to be positive definite, if it simplifies the problem. 
There are in fact two questions:
(1) What is the condition on $H$ that guarantees that such symmetric unitary $U$ exists?
(2) Given that $U$ exists, is it possible to find it using some constructive algorithm?
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Let me mention a few examples for $2\times 2$ matrices. 
Using the Pauli matrices as basis, one easily finds that if $H$ is a symmetric matrix (i.e. linear combination of $\mathbf{1}$, $\sigma_x$ and $\sigma_z$), then the identity matrix $U = \mathbf{1}$ has this property. On the other hand, for $H=\sigma_y$, the choice $U = \sigma_x$ leads to $\sigma_x \cdot \sigma_y \cdot  \sigma_x^\dagger = -\sigma_y = (\sigma_y)^\top$, which solves the problem. 
However, for general matrices and especially of higher dimensions, I found it rather difficult to state anything specific.
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Let me also mention a bit on the physics background: 
One might wonder whether a model for spinless electrons hopping between a collection of sites respects time-reversal symmetry. The time-evolution of the electron states is modelled by a Hermitian Hamiltonian matrix $H$. It follows from more general considerations that spinless time-reversal symmetry is an ($i$) antiunitary symmetry, that ($ii$) commutes with the Hamiltonian, and that ($iii$) squares to $+\mathbf{1}$. 
Since any antiunitary symmetry [condition ($i$)] can be expressed as a unitary followed by complex conjugation, it follows from ($iii$) that the unitary part is symmetric. Then condition ($ii$) then translates into the posed mathematical problem. I am thus trying to check whether the Hamiltonian $H$ exhibits spinless time-reversal symmetry. 
While the symmetry is usually very apparent for a physical motivated basis of the Hilbert space, it could be that the basis in which I am given the Hamiltonian contains complicated phase factors, which make the time-reversal symmetry difficult to reveal.
 A: Edit. The OP has changed the equation to $H^\top=UHU^\ast$. In this case, the equation is always solvable and there is a simple solution. Since $H$ is Hermitian, we can always unitarily diagonalise it as $Q^\ast HQ=D$, where $Q$ is a unitary matrix whose columns form an orthonormal eigenbasis of $H$ and $D$ is a real diagonal matrix. It follows that
$$
Q^\ast HQ=D=\overline{D}=\overline{Q^\ast HQ}=\overline{Q}^\ast H^\top\overline{Q}.
$$
Hence
$$
H^\top=\overline{Q}Q^\ast HQ\overline{Q}^\ast
=\left(\overline{Q}Q^\ast\right)H\left(\overline{Q}Q^\ast\right)^\ast
$$
and one may pick $U=\overline{Q}Q^\ast=\overline{Q}(\overline{Q})^\top$.

(Below was my old answer for the old question, in which the equation concerned was $H^\top=UHU$ for some symmetric unitary matrix $U$.)
This problem looks non-trivial. I haven't any answer but I can make some quick observations:


*

*the desired $U$ exists if and only if there exists a unitary matrix $V$ such that $V^\top HV$ is real;

*a necessary condition for the existence of $U$ is that $-iSK$ is unitarily similar to $iSK$, where $S$ and $K$ denote respectively the real and imaginary parts of $H$ (so that $S$ is real symmetric, $K$ is real skew symmetric and $H=S+iK$).


For the first bullet point, suppose that $H^\top = UHU$ for some symmetric unitary matrix $U$. Being symmetric, $U$ admits a Takagi factorisation $U=V\Sigma V^\top$, where $V$ is unitary and $\Sigma$ is a nonnegative diagonal matrix. As $U$ is unitary, $\Sigma$ is necessarily equal to $I$. Therefore $U=VV^\top$ and $H^\top = UHU$ implies that $V^\ast H^\top \overline{V}=V^\top HV$, i.e. $V^\top HV$ is real. Conversely, if $V^\top HV$ is real, then $H^\top=UHU$ for $U=VV^\top$.
When $n=2$, if $S=QDQ^\top$ is an orthogonal diagonalisation of the symmetric part $S$ of $H$ over $\mathbb R$, one can always pick $V=Q\operatorname{diag}(1,i)$ and $U=V^\top V=Q\operatorname{diag}(1,-1)Q^\top$.
For the second bullet point, suppose $V^\top HV$ is real. Then its symmetric part $V^\top SV$ and skew symmetric part $iV^\top KV$ are real, and so is $V^\top SV\left(\overline{iV^\top KV}\right)=-iV^\top SK\overline{V}$, meaning that $-iSK$ is unitarily similar to a real matrix. By Khakim Ikramov (2010), On complex matrices that are unitarily similar to real matrices, Mathematical Notes, 87(6): 821-827, a complex matrix $A$ is unitarily similar to a real matrix if and only if $A$ and $\overline{A}$ are unitarily similar. Hence $-iSK$ and $iSK$ are necessarily unitarily similar.
In particular, the nonzero eigenvalues of $-iSK$ must occur in sign pairs $(-\lambda,\lambda)$. In view of this, we see that when $n\ge3$, $H^\top=UHU$ more often than not is insolvable. E.g. when $S$ happens to be positive definite, $-iSK$ is similar to the Hermitian matrix $-iS^{1/2}KS^{1/2}$. While the nonzero eigenvalues of $-iK$ do occur in the form of $(-\lambda,\lambda)$ (because $K$ is real skew symmetric), this pairing structure is usually destroyed after a congruence via $S^{1/2}$.
