Although the term 'high degree of symmetry' is somewhat subjective the only molecules with dipoles are those that belong to the point groups $C_1$ (no symmetry elements other than rotation by 360 degrees), the group $C_s$ which has only one mirror plane and the groups $C_n$ ($n \ge 2$ ) which have rotation axes of $2\pi/n$ degrees and $C_{nv}$ which in addition have a (vertical) mirror plane along a rotation axis. An example of a $C_{2v}$ molecule is $\ce{H2O}$, and $\ce{NH3}$ is $C_{3v}$. All these point groups do perhaps have 'lower' symmetry than say an octahedral group and so the question suggests that molecules with low symmetry have a dipole moment and the rest don't. (You can see 3D models of many molecules of all common point groups at molecule-viewer.com and add/remove their symmetry elements)
As @Dan Burden suggest the question is therefore most probably to do with the presence or absence of a dipole moment. One might suppose that a dipole will cause the molecule to have a larger van der waals interaction energy due to dipole-dipole interaction, than a similar molecule without such a dipole and so have higher boiling and melting points. Isomers for example, could be used to test this.
Three such could be 1,2-, 1,3- and 1,4-dichlorobenzene; I chose at random. The 1,2- and 1,3- have a dipole ($C_{2v}$ point group) but the 1,4- does not ($D_{2h}$ point group). The data (from Wikipedia pages) are below
boiling point (centigrade) $1,2 = 180.5, ~~~~ 1,3 = 172,~~~~ 1,4 = 174 $
so not much difference
melting point(centigrade) $1,2 = -17, ~~~~1,3= -22, ~~~~ 1,4 = 53$
not what one would expect if dipole interaction is important, quite the opposite in fact.
vapour pressure $1,2= 1.0, ~~~~ 1,4 = 1.3$ mmHg so again, not much difference here
To me the question does not have clear answers and it seems possible to choose molecules to support whatever the 'official' answers are or to show the opposite.