We have a permutation $6,2,3,4,5,1$ of $1,2,3,4,5,6$.

Denote $V$ the set of dwarves whose both neighbors remained and let $W$ denote the set of dwarves whose exactly one of the neighbors is a member of $V$, but not itself. We must have $|V|=3$ and $|W|\leq 3$ It's not hard to conclude that there exist two consecutive dwarves in $V$, and then three consecutive dwarves. We get that $|W|=2$. Then note that the other dwarf whose only one of the neighbors remained must not be a neighbor of any member of $W$. So among $4$ dwarves other than that in $V\cup W$, there exist exactly one "directed" neighbor relationships, which is an odd number. Contradiction.