How to distingush the $\Delta$ circuit from the $Y$ circuit by external measurement? We know how to calculate the parameters of a $Y$ circuit in order to get its $\Delta$ counterpart, and vice versa, as well.
My question is as follows.
Assume that we have an equivalent pair of a $\Delta$ and a $Y$ circuit both locked in boxes.

Is there a possibility two reveal which box contains which circuit if we have access only to the outside terminals?

EDIT
The author of the first comment asks the question:
"Doesn't the word "equivalent" imply that you cannot distinguish between the two circuits?"
Yes, if we stay within the realm of the model. But in reality? 
Assume now, that we have a chance to do measurements on a set of impedances.  Having done some experiments with the possible parts, we surrender them and a secret agent assembles them into a $Y$ or $\Delta$ circuit.
What would we have to do if our task is to find out what the secret agent did?

 A: 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.
A: Maybe we could add reactive loads with the resistive load(in case the load is purely resistive) then connectveach to a three phase supply and obserce for the differenr harmonics?
