For signature $s= \{ f\}, n_f=1$, show that if for structures $M, N$, $f^N, f^M$ have only one orbit, of infinite order, then $M,N$ are isomorphic

Question

For signature $s= \{ f\}, n_f=1$, show that if for structures $M, N$, $f^N, f^M$ have only one orbit, of infinite order, then $M,N$ are isomorphic.

So we know that for some $m \in M$, the set $\{f^k(m): k \in \mathbb Z \}$ is infinite, and we know that for some $n \in N$, the set $\{f^k(n): k \in \mathbb Z \}$ is infinite (the first $f$ is $f^M$, the second one is $f^N$).

My initial guess is that $\alpha ((f^M)^k(m))=(f^N)^k(n)$ will suffice as an isomorphism. However I cant really think of a way to show it's surjective.

Note: The question defines the order of $a \in M$ as the set $\{f^k(a): k\in \mathbb Z \}$, and when $k \lt 0, f^k(a)=(f^{-1}\circ f^{-1}\circ \cdots \circ f^{-1})(a)$, $k$ times.

Thank you in advance!