Is Ising model on any infinite $2k$-regular graph (where the vertex degree is exactly $2k$) equal to Ising model on $\mathbb{Z}^k$ ($\mathbb{Z}^k$ lattice) ( where the vertex degree is $2k$ as well but in special way: only vertices with distance one are connected to each other) ?
Maybe if we could say any infinite $2k$-regular graph is isomorphic to $\mathbb{Z}^k$lattice, the answer would be yes. Are they isomorphic?
the answer is no. First of all, notice that a $2k$-regular tree is not isomorphic to $\mathbb{Z}^k$, since the first graph does not have cycles
while the second one does.
Beyond the absence of the isomorphism between these two graphs (in many reasonable senses of equivalence) the Ising model on both the regular tree and the hypercubic lattice ($\mathbb{Z}^d$) are not equivalent. For example, if we think about the equivalence as being the number of Gibbs measures, the first neighbors ferromagnetic Ising model with constant magnetic field $h\in\mathbb{R}$ provides an example of non-equivalence.
Indeed, this model in the hypercubic lattice $\mathbb{Z}^k$, has no first order phase transition, due to the Lee-Yang theorem and therefore for any inverse temperature $\beta>0$ the set of the Gibbs measures is a singleton. On the other hand, analogous model on $2k$-regular tree has at least two Gibbs measures for any inverse tempertature $\beta>0$ if $|h|$ is small enough, due to a theorem of Jonasson and Steif, see:
Jonasson, J. and Steif, J.: Amenability and Phase Transition in the Ising Model. Journal of Theoretical Probability 12 , 549-559, 1999.
I should add that, roughly speaking, the Ising model on a regular tree can be thought as a limit of the Ising models on the $k$-dimensional hypercubic lattice, when $k\to\infty$.
