The $Y$-$\Delta$ transform assumes linear circuit elements. Real circuit elements are nonlinear, so it should be possible to distinguish between the two cases by using sufficiently large voltages/currents. The current flow is "spread out" over more resistors in the $\Delta$ circuit than in the $Y$ circuit (think about connecting only two of the terminals, for instance), so the nonlinear responses will be different.
Edited to add an example:
Just to illustrate the idea, imagine that all the $Y$ resistors are identical ($R_1 = R_2 = R_3 = R_Y$), but that $R_Y$ changes value when the current $I$ through the resistor passes some threshold: $R_Y = R_Y^0$ for $I < I_Y$ and $R_Y = R_Y^1$ for $I > I_Y$. You can imagine the change happening smoothly over some range around $I_Y$, if you wish.
By symmetry, the $\Delta$ resistors must also be identical, $R_a = R_b = R_c = R_{\Delta}$, and must change value when the current passes some threshold: $R_{\Delta} = R_{\Delta}^0$ for $I < I_{\Delta}$ and $R_{\Delta} = R_{\Delta}^1$ for $I > I_{\Delta}$. However, the secret agent is free to choose $R_{\Delta}^0, R_{\Delta}^1$ and $I_{\Delta}$ as he or she wishes.
Equivalence of the circuits at low currents forces $R_{\Delta}^0 = 3 R_Y^0$. Equivalence at very high currents forces $R_{\Delta}^1 = 3 R_Y^1$. I claim that we can distinguish between the two circuits by doing a two-terminal measurement (say across nodes 1 and 2), and ramping the applied voltage/current:
For the $Y$ circuit, the same current flows through resistors $R_1$ and $R_2$. These will change value at the same current, so the IV curve will change slope only once. For the $\Delta$ circuit, on the other hand, more current flows though $R_c$ than through $R_a$ and $R_b$. Therefore, $R_c$ will change value before $R_a$ and $R_b$. Consequently, the IV curve will change slope twice.
