How to construct a NFA that the third symbol and the third from last symbol are same? Suppose the alphabet is {a,b}. I have tried coming up with NFA1: third symbol is a, and NFA2: third from last symbol is a. Then NFA3 and NFA4 similar but with symbol of b. Then I use intersection of NFA1, NFA2, and intersection of NFA3, NFA4, and add $\epsilon$ transition. But things seem to go wrong here. When doing the intersection of, say, NFA1, and NFA2, obviously the new NFA would not accept strings such as aba, aaa. 

I have been trying different methods but still stuck here for hours. Any help of hint would be much appreciated!
Hint:

Here is a NFA that accepts exactly those strings whose first and last character are both $a$:



Can you modify and extend this to become a solution to your problem?