Under what conditions is the star-mesh transform invertible?
For the mesh-star conversion the problem is that you have more equations than variables, so the number of bond \$N_b\$ is: \$N_b=N_e-N_v\$, where \$N_e\$ is the number of equations, equal also to the number of the resistance in the mesh, \$N_v\$ is the number of variables, equal to the number of the resistance in the star. In the 4-case, I've demonstrated that the bonds for the transformation are \$G_{AB}G_{CD}=G_{AC}G_{BD}=G_{AD}G_{BC}\$, in other words the products between the resistance without node in common must be the same.

Ps: The "demonstration" is:
The formula for the star-mesh transformation is \$G_{XY}=\frac{G_XG_Y}{G_{TOT}}\$ with \$G_{TOT}=\sum_{i=1}^nG_i\$. So, assuming \$G_{XY}\ne0\$, we can divide two of those equation and obtain \$\frac{G_X}{G_Y}=\frac{G_{XZ}}{G_{YZ}}\$ for every Z different from X or Y. In the 4-case this means 6 equations, one is the following: \$\frac{G_A}{G_B}=\frac{G_{AC}}{G_{BC}}=\frac{G_{AD}}{G_{BD}}\Rightarrow G_{AC}G_{BD}=G_{AD}G_{BC}\$. We get the same result from: \$\frac{G_C}{G_D}=\frac{G_{AC}}{G_{AD}}=\frac{G_{BC}}{G_{BD}}\$. From the last 4 equations we obtain \$G_{AB}G_{CD}=G_{AD}G_{BC}\$ and \$G_{AB}G_{CD}=G_{AC}G_{BD}\$ and so we finally have the \$G_{AB}G_{CD}=G_{AC}G_{BD}=G_{AD}G_{BC}\$ condition. So this is a necessary condition. But if the ratio between any two conductances of the mesh is known ,we can express the \$G_{TOT}\$ depending on only one of those, like \$G_{TOT}=G_{A}+G_B+G_C+G_D=G_A(1+\beta+\gamma+\delta)\$, where \$\beta=\frac{G_B}{G_A}=\frac{G_{BC}}{G_{AC}}=\frac{G_{BD}}{G_{AD}}\$, and so on..\$\Rightarrow G_{AB}=\frac{G_AG_B}{G_{TOT}}=\frac{G_AG_B}{G_A(1+\beta+\gamma+\delta)}=\frac{G_B}{(1+\beta+\gamma+\delta)}\Rightarrow G_B=G_{AB}(1+\beta+\gamma+\delta)\$. With similar calculations we can find all the 4 conductances(resistances) of the star. 

I suppose all of this means the condition is also a sufficient condition.