characteristic polynomial of skew Hermitian matrix coefficients are real??

Let $$X$$ be a skew Hermitian matrix.

Is it true that its characteristic polynomial of $$X$$ i.e., $$\det(\lambda I-X)$$ has real coefficients?

Consider the matrix $$X=\begin{bmatrix} -i&2+i\\-2+i &0\end{bmatrix}$$. This is skew-Hermitian. Its characteristic polynomial is $$\lambda^2+\lambda i+5$$. So, characteristic polynomial of skew-Hermitian matrix need not have real coefficients.

Consider the matrix $$\frac{1}{i}X=\frac{1}{i}\begin{bmatrix} -i&2+i\\-2+i &0\end{bmatrix}$$. Its characteristic polynomial is $$\lambda^2+\lambda-5$$ whose coefficintes are real.

I have checked some random examples and it turns out that for all of them ($$X$$ is skew-Hermitian), the characteristic polynomial of $$\frac{1}{i}X$$ is with real coefficients. Is this true in general? I think it is true. Can not think of a proof in general. Any suggestions are welcome.

The statement is

If $$X$$ is skew-Hermitian, characteristic polynomial of $$\frac{1}{i}X$$ is with real coefficients.

One thing is clear. As skew-Hermitian has Eigenvalues purely imaginary, trace is $$0$$ or $$ai$$ for some $$a\in \mathbb{R}$$. So, trace of $$\frac{1}{i}X$$ is $$\frac{1}{i}(ai)=a$$ i.e., real. So, one coefficient of characteristic polynomial is real. I can not think of general proof for other coefficients.

This is true. We have that $$\frac{1}{i}=-i$$, and if $$X$$ is skew-Hermitian, $$-iX$$ is Hermitian. To check this, denoting $$X^H$$ to be the conjugate transpose, we have $$(-iX)^H=\overline{-i}(X^H)=i(-X)=-iX$$ Now, Hermitian matrices have all real eigenvalues, so the characteristic equation will also have all real coefficients.
Another way to see that all the eigenvalues of $$-iX$$ are real is to use the fact that skew-Hermitian matrices are diagonalizable, and move the $$-i$$ coefficient to the diagonal matrix, whose entries are the eigenvalues of $$X$$.