Is a complete circuit required for capacitors to redistribute charges? Suppose I have two capacitors of different capacitance $C_1$, $C_2$and they have been charged prior connecting by voltage of $V_1$ and $V_2$ respectively the positive plate of one capacitor is connected to negative plate of the other.The other plates are not connected , we thus have an incomplete loop.
Had it been a complete loop , I would have easily redistributed charges . The two plates which have been connected are at different potentials , I'm interested to know what happens next , what changes will occur in the wire and all the plates , will a change be even there or not ? If there is no change then what happens to free electrons in wire why won't they move provided the wire's end are at different potentials ? 
Please help , I'm unable to think around this.

Here is what I think I going to happen , I've taken a case where capacitors are charged with 3 microcoulomb and 2 microcoulomb and then their plates are joined as shown , the system is now 3conductors - leftmost plate , the two inner plates connected taken as whole and rightmost plate and therefore charges in beginning and end on them will be same but there will be redistribution as shown in my diagram which will ensure a system of lowest energy , because obviously the two inner plates won't just stand when there are different charges on them and they are connected . 
if it suites you then assume that both capacitors were identical .
 A: 
The two plates which have been connected are at different potentials , I'm interested to know what happens next , what changes will occur in the wire and all the plates , will a change be even there or not ?

Once you disconnect the individual capacitors from their respective voltage sources they no longer have a well-defined potential. They are "floating". The potential difference is fixed by the original voltage sources, but the absolute potential is not (absolute potential is arbitrary anyway).
So what happens is that when the two wires are connected then the potentials of the two connected plates becomes the same. There is no redistribution of charges (neglecting any small parasitic or self capacitance of the wire) because there is no need for it. If you ground any of the three points then the voltages are all well-defined, but until then the voltages are free to float as needed, with only the voltage difference across the plates being fixed. There is no need for them to redistribute charges to become equipotential. 
EDIT (to address new content in the question):
The thing that you have labeled “potential difference” does not exist. There is a potential difference across each capacitor established by the sources, but when they are disconnected from the sources then the voltages are floating. When the wires touch they are already equipotential, having both “floated” to the ambient potential. 

there will be redistribution as shown in my diagram which will ensure a system of lowest energy

Any redistribution will raise the energy of the system. With the charges on each capacitor as they originally are the E field is concentrated in the small volume between the plates. The field lines leaving the positive plate are very short and terminate on the negative plate. As you move charges away from the plate some of the field lines will become much longer. You will therefore increase the volume occupied by the field and thereby increase the energy. 
A: When you remove the battery there will be nowhere for the charge at the lead ends to go, so it will stay the same. As there is no redistribution on the lead ends, by symmetry there will be no redistribution internally.
You can also thinking of it as the capacitors charging up while current is flowing. Once equilibrium is reached the current has stopped and the voltage across the capacitor matches that of the battery. With the removal of the battery, nothing has changed from this equilibrium state.
A: The charges on the unconnected plates have no where to go. The key point is the charges on the other plate of each capacitor are bound to those plates by the electrostatic force of the unconnected plates.
Bottom line: In my opinion, nothing happens. You simply wind up with two capacitors in series with a voltage of $V_{1}$ on one and $V_{2}$ on the other. Total charge remains the same (conservation of charge). It doesn't matter if the voltages were originally equal or not, or if the original charge on each were equal or not. 
I am amenable to change my opinion given sound technical arguments to the contrary.
I would have to add, however, that if after connecting the plates together you then connect the open plates together, than charge can redistribute themselves causing the voltages to change (one capacitor charging another). Because for a complete circuit with capacitors in series, the charge on each capacitor will be the same.
Hope this helps.
