For what it's worth, in the case when $U,V$, and $W$ all have dimension $2$ (i.e., a case that is much simpler than the $4$-dimensional one you're interested in), it is known that there are exactly six orbits. In particular, every vector is in the orbit of exactly one of these six vectors (where $\{\mathbf{e}_1,\mathbf{e}_2\}$ is some fixed basis of $U,V,W$):

 1. $\mathbf{e}_1 \otimes \mathbf{e}_1 \otimes \mathbf{e}_1$
 2. $\mathbf{e}_1 \otimes \mathbf{e}_1 \otimes \mathbf{e}_1 + \mathbf{e}_1 \otimes \mathbf{e}_2 \otimes \mathbf{e}_2$
 3. $\mathbf{e}_1 \otimes \mathbf{e}_1 \otimes \mathbf{e}_1 + \mathbf{e}_2 \otimes \mathbf{e}_1 \otimes \mathbf{e}_2$
 4. $\mathbf{e}_1 \otimes \mathbf{e}_1 \otimes \mathbf{e}_1 + \mathbf{e}_2 \otimes \mathbf{e}_2 \otimes \mathbf{e}_1$
 5. $\mathbf{e}_1 \otimes \mathbf{e}_1 \otimes \mathbf{e}_1 + \mathbf{e}_2 \otimes \mathbf{e}_2 \otimes \mathbf{e}_2$
 6. $\mathbf{e}_1 \otimes \mathbf{e}_1 \otimes \mathbf{e}_2 + \mathbf{e}_1 \otimes \mathbf{e}_2 \otimes \mathbf{e}_1 + \mathbf{e}_2 \otimes \mathbf{e}_1 \otimes \mathbf{e}_1$

Furthermore, a generic vector in $U \otimes V \otimes W$ belongs to the orbit of the vector 5 above: the other orbits all have measure $0$.