Is the following equality correct?
$$H^{\otimes 2} \, \mathsf{CNOT} \, H^{\otimes 2} |xy\rangle = \mathsf{CNOT} |yx\rangle$$
The solution says yes. It claims that the LHS maps $|10\rangle$ to $|10\rangle$. But isn't it the case that $\mathsf{CNOT} |11\rangle$ rather than $\mathsf{CNOT} |01\rangle$ equals $|10\rangle$? If yes, then why is this equality correct?
You are correct, and your solutions are wrong. Acting on the basis ket $|10 \rangle$, the right-hand side maps to \begin{align} \mathsf{CNOT} \ \mathsf{SWAP}|10\rangle & = \mathsf{CNOT}|01\rangle \\ & = |01\rangle. \end{align} Their assertion for the left-hand side is correct, though: \begin{align} H^{\otimes 2} \, \mathsf{CNOT} \, H^{\otimes 2}|10\rangle & = H^{\otimes 2} \, \mathsf{CNOT} \frac{|0\rangle-|1\rangle}{\sqrt{2}}|+\rangle \\& = H^{\otimes 2} \frac{|0\rangle|+\rangle-|1\rangle\,X|+\rangle}{\sqrt{2}} \\& = H^{\otimes 2} \frac{|0\rangle|+\rangle-|1\rangle|+\rangle}{\sqrt{2}} \\& = H^{\otimes 2} \frac{|0\rangle-|1\rangle}{\sqrt{2}}|+\rangle \\ & = |10\rangle. \end{align} As such, under the standard conventions, the equality is false.
On the other hand, it is true that $$ H^{\otimes 2} \, \mathsf{CNOT} \, H^{\otimes 2} = \mathsf{SWAP} \ \mathsf{CNOT} \ \mathsf{SWAP}. $$ Either way, you should check with your instructor (or, if self-studying, find a better problem set).
