Let $f(n)$ be the number of different sequences of plates at the end. Then $f(2)=1$. $f(n+1)=2f(n)$. Why? If we did the $n,n+1$ move at the beginning then there would be $f(n)$ different outcomes since it would be the same as having $n$ houses, only the nth house would say n+1 for some weird reason. In fact if we do the $n,n+1$ move before doing the n-1,n move we will always get one of the $f(n)$ combinations. The other case is doing the n,n+1 move after doing the n-1,n move. Note that if we ignored the $n$, and $n+1$ plates it would look exactly like one of the $f(n)$ combinations. and we know that in this case house $n$ will have the $n+1$ plate and the house $n+1$ wil have the plate that would have been in house $n$ in the $f(n)$ configuration. Therefore $f(n+1)=2f(n)$. and knowing $f(2)=1$ we conclude $f(n)=2^{n-1}$