Question: > 
> The plates of an isolated parallel plate capacitor with a
>  capacitance C carry a charge Q. The plate separation is d.
>  Initially, the space between the plates contains only air.
>  Then, an isolated metal sheet of thickness 0.5d is inserted
>  between, but not touching, the plates. How does the
>  potential difference between the plates change as a result
>  of inserting the metal sheet?
> 
> 
> 

The answer is: The potential difference will decrease. How do I get this result, it seems right but how do I reason it out? 

$$V = Ed$$

There was mention about "*E is the same outside the conductor. But
the field inside is 0.*". But in the formula, E is the field outside conductor right? If so it remains the same? Then $d = 0.5d$?

Answer: 
If there is a charge $Q$ and $-Q$ on each plate of the capacitor , when you insert a perfect conductor between the plates (parallel), you simply will have a charge $+Q$ on one side(facing negative plate of capacitor) and $-Q$ on other side of the inserted plate. (Given that the area of plates is large enough to assume constant electric field between them) 

Now , because the field between two (effectively) infinite plates doesn't depend on their distance, you will have the same field as before (without the new conductor).So , as you mentioned, the new potential difference is $E(d\_0-d\_{metal})$ .

Remember that according to uniqueness theorems in electrostatics, when you find an answer for a problem (that is consistent with equations) , this is the **correct and only possible** answer . So, in electrostatics problems you should try to find a *story* to create the final form of the system from a simple one , through which you can find the answer more easily.
Yes d=0.5d, Basically start thinking in terms of work per unit charge. So work you do on $1$ coulomb is $E$x$0.5d$. as field in conductor is $0$ in electrostatic condition , you don't have to do any work at all in that region . so $\Delta V$ is 0 in that region . Hence , $\Delta V\_{net}= E\_{outside}$x$0.5d$. = V for the capacitor . And calculate the new field in the two new air regions divided by the metal plate , you will find that by principle of superposition , it remains same . So it is because of this calculation that Net field also comes out to be same as before, that you're able to use E , basically think in terms of work and everything will be resolved . And then you will know , you will have to evaluate E everywhere again because now the metal plate has been polarised .