For matrices $A$ it is well known that the spectrum is invariant under transpose $\sigma(A^T) = \sigma(A)$. Furthermore, the spectrum of the adjoint matrix $\sigma(A^*) = \overline{ \sigma(A)}$ the complex conjugation of the spectrum of the matrix $\sigma(A)$. The results also has immediate generalisations to operators on Hilbert or Banach spaces using the Hilbert or Banach space adjoints respectively.

Suppose now, that we have an operator on the space of trace class operators $TC( \mathcal{H})$ for some Hilbert space $\mathcal{H}$ with a given basis $\vert i \rangle_{i \in \mathbb{N}}$. Then $TC( \mathcal{H})$ is spanned by the elementary matrices $ E_{ij} = \vert i \rangle \langle j \vert_{i,j \in \mathbb{N}}$ closed in the trace norm. Thus, I can describe any operator $A \in \mathcal{B}( TC( \mathcal{H})) $ in terms of its matrix elements \begin{align*} A_{(i,j), (k,l)} = \langle k \mid, A \left( \vert i \rangle \langle j \vert \right) l \rangle. \end{align*} This allows me to define a "transpose" $\tilde A$ operator with respect to our choice of basis by defining it on matrix elements \begin{align*} \langle k \mid, \tilde A \left( \vert i \rangle \langle j \vert \right) l \rangle = A_{ (k,l),(i,j)} \end{align*} and extending it by linearity. I would expect that $\tilde A \in \mathcal{B}( TC( \mathcal{H}))$ and that $\sigma_{ \mathcal{B}( TC( \mathcal{H}))}(\tilde A) = \sigma_{ \mathcal{B}( TC( \mathcal{H}))}(A)$, but how would I go about proving such a statement?

In other words: If I have a trace-class operator with matrix elements $A_{(i,j), (k,l)}$ and "transpose" the operator so that it has matrix elements $A_{(k,l),(i,j)} $ is the operator then still bounded operator on trace class operators?