Consider the Dirac operator
$$ H = \begin{pmatrix} m & -i\partial_z \\ -i\partial_{\bar z} & -m \end{pmatrix},$$ where $\partial_{\bar z}$ is the Cauchy-Riemann operator and $m \ge 0.$
It is not hard to see that the spectrum of this operator is symmetric with respect to zero.
However, does there exist a simple unitary $T$ such that
$$THT^*=-H?$$
If $m$ was zero, then already $$T = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$ would do, what about this general case?
With $z=x+iy$, we use the Fourier transformation in $(x,y)$ to see that $H$ is unitarily equivalent to $$ \frac12\begin{pmatrix}2m&\xi-i\eta\\ \xi+i\eta&-2m\end{pmatrix}, \text{whose eigenvalues are } \lambda_\pm=\pm\sqrt{m^2+\frac{\vert\zeta\vert^2}{4}}, \ \zeta=\xi+i\eta. $$ With $\mu=\sqrt{m^2+\frac{\vert\zeta\vert^2}{4}},$ we have $$ \begin{pmatrix}0&1\\ -1&0\end{pmatrix}\begin{pmatrix}\mu&0\\ 0&-\mu\end{pmatrix}\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}= \begin{pmatrix}-\mu&0\\ 0&\mu\end{pmatrix}, $$ so that $H$ is indeed unitarily equivalent to $-H$. All calculations can be made explicitly.
It may be of interest to note that the $\pm E$ symmetry of the spectrum of the Dirac operator holds not only for a constant $m$, but also for a spatially dependent $m(x,y)$,
$$H = \begin{pmatrix} m(x,y) & -i\partial_x-\partial_y \\ -i\partial_x+\partial_y & -m(x,y) \end{pmatrix}.$$
This follows from the anti-unitary symmetry
$$T\bar{H}T^\ast=-H,\;\;\text{with}\;\;T = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$
The overline $\bar{H}$ denotes complex conjugation.
Then if $E$ is a (real) eigenvalue of $H$, we have $$0=\det(E-H)=\det(E-\bar{H})=\det(E+T^\ast HT)=\det(E+H),$$ so $-E$ is also an eigenvalue.
In physics this is called particle-hole symmetry (or charge-conjugation symmetry), the matrix $T$ exchanges an electron state $(1,0)$ into a hole state $(0,1)$.
