Yes, it's true. Let's enumerate the seats from left to right $1,2,\ldots,n$. Suppose the spectator, with the ticket No $1$, is placed in the $k$-th seat. We will prove that it is possible to bring that spectator to seat $1$, complying the rules, with the final configuration being also derangement except the first seat. Thus, consecutively, after final number of moves, we will do the job. If the ticket $k$ is not placed in the $k-1$-th seat, we can exchange these spectators moving ticket $1$ in the desired direction. If not, suppose that $\ell\,,\, \ell < k$ is the least seat with the property: " the ticket $i+1$ is placed in the $i$-th seat, $i=\ell,\ell+1,\ldots, k-1$ ". If $\ell=1$ we move the ticket $1$ (with the spectator which holds it )consecutively to the left, till it is placed in the first seat. After that, in the seats $1,2,\ldots, k$ are the right spectators. If $\ell>1$, we first move consecutively to the right the spectator in the $\ell-1$-th seat till he reaches the $k$-th seat and then we move the spectator with ticket $1$ (which at that moment seats in the $k-1$-th seat) to the left till he reaches the $\ell-1$-th seat. In this way we are at position, which is also derangement, and ticket $1$ is more to the left than it was initially placed. Thus, doing this trick, after some time, we will place the ticket $1$ in the first seat.