homework and exercises - Uhlmann's Theorem: proof of $text{tr}(A^{dagger} B) = langle m | A otimes B |mrangle $

In p228, Chapter 9 of Mark Wilde's text , in the course of proving Uhlmann's theorem for quantum fidelity, it claims $$\sum_{i,j} \langle i|^R \langle i|^A (U^R \otimes (\sqrt{\rho}\sqrt{\sigma})^A) |j\rangle^R |j\rangle^A $$ $$=\sum_{i,j} \langle i|^R \langle i|^A (I^R \otimes (\sqrt{\rho}\sqrt{\sigma}U^T)^A) |j\rangle^R |j\rangle^A $$ which are equations (9.97) and (9.98) in the aforementioned text.

Meanwhile, in Nielsen & Chuang's text, exercise 9.16 requires to prove that $$\text{tr}(A^{\dagger} B) = \langle m | A \otimes B |m\rangle $$ for $|m\rangle = \sum_{i} |i\rangle|i\rangle $ where $\{ |i\rangle \}$ is an orthonormal basis on some Hilbert space and A and B are operators on that space.

Each thing above is crucial in proof of Uhlmann theorem in respective textbook but I have no idea why they hold. $\text{tr}(A^{\dagger} B) = \sum_{i,j} {a_{ij}}^{*}b_{ij}$ whereas $ \langle m | A \otimes B |m\rangle = \sum_{i,j} {a_{ij}} b_{ij } $ so why they equal? Could anybody give me any hint?