No, NP Completeness states that you can go either way.
Specifically: A problem X is NP Complete if:
1: X is in NP.
2: Every problem in P can be reduced to X in polynomial time.
We have established the first condition by intuition-- pathfinding is easy.
Immediately, we know, then, that there exists some way to reduce this to 3-SAT, in a way general for any level.
Reducing 3-SAT to this, however, proves the second criterion (because 3-SAT is NP-complete, so by doing two reductions you can reduce any NP problem to this, in time at worst the cost of doing sum of the two polynomials describing the complexity of the reductions).
So we can also convert 3-SAT-- or any NP-complete problem, in fact, and even more generally, any NP problem-- to this in poly-time.