Well you want kx+b = x (mod m) to have a solution for all b and m (here b depends on how much earlier the hare enters the cycle). This means you'll need to have k-1 comprime to m for all m>0. Pretty sure k-1 = 1 is the only solution.
As a counter example where both k and k-1 are prime consider k=3. Then:
As a counter example where both k and k-1 are prime consider k=3. Then:
3 x + b = x (mod 2)
will fail to have a solution for odd b.