If a matrix as well as its Hermitian part both have determinant one, must the matrix be Hermitian? If $x\in\mathrm{M}_2(\mathbb{C})$,
$y=\dfrac{x+x^{\dagger}}{2}$, and $z=\dfrac{z-z^{\dagger}}{2}$,
then $x=y+z$.
Also, $y$ and $z$ respectively are Hermitian and anti-Hermitian, i.e. $y^{\dagger}=y$ and $z^{\dagger}=-z$, where $^\dagger$ denotes the conjugate transpose.
Now suppose $\det(x)=\det(y)=1$.
Does this force $x^{\dagger}=x$?
I cannot find a counterexample but I can't prove it either.
Some thoughts:
Let $H$ and $A$ respectively denote the sets of $2\times2$ complex Hermitian and anti-Hermitian matrices, and
consider the map $\phi:\mathrm{M}_2(\mathbb{C})\rightarrow H\times A$
which takes $x$ as above to $(y,z)$.
Then $x\in H\iff\phi(x)=(x,0)$.
Also, $\phi$ is a vector space isomorphism if we think of $\mathrm{M}_2(\mathbb{\mathbb{C}})$
as an 8 dimensional real vector space, so taking the first coordinate of its image gives a projection onto $H$ as a 4 dimensional real subspace.  I want to say that this projection cannot fix the determinant of $x$ unless $x$ is already in its image, but it's difficult to say anything in particular about the determinant here because it maps into $\mathbb{C}$ and the vector spaces I'm thinking about are real.
 A: We have $x = y + z$. Any Hermitian matrix is orthogonally diagonalizable with real eigenvalues, so (after conjugating by a unitary matrix) we can assume $y$ is a diagonal matrix. (Note that conjugating by a unitary matrix will preserve the Hermitian/anti-Hermitian qualities of $y$ and $z$, as well as the determinants of each). By the determinant constraint we have:
$$y = \begin{bmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{bmatrix}$$
Where $\lambda \in \mathbb{R}^{\times}$ is nonzero. We also have:
$$z = \begin{bmatrix} ia & w \\ -\overline{w} & ib \end{bmatrix}$$
Here $a,b \in \mathbb{R}$ and $w \in \mathbb{C}$. So $x$ has determinant $1$ if and only if:
$$(ia + \lambda)(ib + \lambda^{-1}) - (w)(-\overline{w}) = 1$$
$$-ab + i(a\lambda^{-1} + b\lambda) + 1 + w\overline{w} = 1$$
$$(w\overline{w} - ab) = 0 \qquad \text{and} \qquad a\lambda^{-1} + b\lambda = 0$$
The last line comes from separating the real and imaginary parts of the above equation. Now the second equation gives:
$$a = -b\lambda^{2}$$
And so:
$$w\overline{w} + b^{2}\lambda^{2} = 0$$
The last line is the sum of two nonnegative real numbers, which is zero if and only if both its terms are zero. So we get $w = 0$ and $b = 0$, it follows that $a = 0$ and so $z = 0$ and $x = y$ as conjectured.
This solves your question is 2 dimensions. I'm not sure about higher dimensions. This method will work in practice but it will quickly become burdensome, so if it is true in general then probably there is a nicer proof.
