According to the answer by joshphysics, in the first example he gave, he mentioned that the state space of 1D free particle is isomorphic to that of 3D. My understanding of this is: both spaces are infinite dimensional, yet the first space has DOF of 1, and the second, 3. Should both spaces be finite dimensional as in the case of spin, the claim wouldn't hold anymore, right?
Now moving to udrv's answer here to another question,
Although you can view x and y as labels for a 2D particle, there is also an isomorphism with a system of two distinguishable 1D particles. Similarly, a system of two distinguishable 3D particles is isomorphic to one particle living in a 6D configuration space.
But isn't a system of two distinguishable 1D particles also isomorphic two 3D particles and thus also one 6D particles? While this could be easily explained by the isomorphism among all infinite dimensional Hilbert spaces, I'd like to reassure that beyond isomorphism, the system of one 2D particle is one single Hilbert space and not the direct product of two Hilbert spaces.
In a general conclusion, may I say that a state space with DOF n, is always a direct product of n other Hilbert spaces, each of which may or may not be finite dimensional? (i.e. it doesn't really have to do with the dimension of the total Hilbert space directly, but it has to do with the number of "factors" in the direct product, which in turn results in the total Hilbert space, whose dimension is the product of all the "factor" dimensions)