Let the two matrices $H$ and $H'$ be defined as follows:
$$H = M(k) \sigma_0 \otimes \sigma_z + \lambda (\sin k_x \sigma_x \otimes \sigma_x + \sin k_y \sigma_y \otimes \sigma_x + \sin k_z \sigma_z \otimes \sigma_x)$$
and
$$H' = M(k) \sigma_z \otimes \sigma_0 + \lambda (\sin k_x \sigma_x \otimes \sigma_x + \sin k_y \sigma_x \otimes \sigma_y + \sin k_z \sigma_x \otimes \sigma_z),$$
where $\sigma_i$ are the Pauli matrices:
$$ \sigma_x = \begin{pmatrix} 0 & 1 \\\ 1 & 0 \end{pmatrix}, \sigma_y = \begin{pmatrix} 0 & -i \\\ i & 0 \end{pmatrix}, \sigma_z =\begin{pmatrix} 1 & 0 \\\ 0 & -1 \end{pmatrix},$$
and $\sigma_0$ is the $2 \times 2$ identity matrix. I want to find a unitary transformation $U$ such that $H' = U^{\dagger} H U$. I can write the matrices explicitly and try to solve a system of equations, but what I am really after is a more direct way to do it, especially noting that the difference between $H$ and $H'$ is just switching the order of the Kroenecker products betwen the $\sigma_i$ and $\sigma_i$.
The unitary $U$ you are looking for is the flip $$ F=\begin{pmatrix} 1&0&0&0\\ 0&0&1&0\\ 0&1&0&0\\ 0&0&0&1 \end{pmatrix}. $$ Because $F$ is a permutation matrix it in particular is unitary. As is readily verified this is the (unique) matrix $F\in\mathbb C^{4\times 4}$ which satisfies $F( x\otimes y)= y\otimes x$ for all $ x, y\in\mathbb C^2$ -- hence why the name "flip".
The reason why this is a solution to the equation $H'=U^\dagger HU$ you are interested in is the fact that $$ F^\dagger(A\otimes B)F=F(A\otimes B)F=B\otimes A\tag{1} $$ for all $A,B\in\mathbb C^{2\times 2}$ (note $F=F^\dagger$). As you rightfully observed any matrix which satisfies (1) transforms $H'$ into $H$ (and vice versa).
To verify (1) note that it suffices to prove this for rank-1 matrices: on the one hand (1) is linear in both $A$ and $B$, and on the other hand every matrix can be written as a linear combination of rank-1 operators ($A=\sum_{i,j}\langle e_i,Ae_j\rangle|e_i\rangle\langle e_j|$). Now given $ x_1, x_2, y_1, y_2\in\mathbb C^2$ the flip-property of $F$ implies \begin{align*} F(| x_1\rangle\langle x_2|\otimes| y_1\rangle\langle y_2|)F&=F| x_1\otimes y_1\rangle\langle x_2\otimes y_2|F\\ &=| y_1\otimes x_1\rangle\langle y_2\otimes x_2|=| y_1\rangle\langle y_2|\otimes| x_1\rangle\langle x_2| \end{align*} so (1) holds for all rank-1 matrices as desired.