Let $d = (n, k)$. If $d > 1$ and $n = dm$, $k = dl$, then by performing the global move
$$e_1 e_2 \dots e_k e_{k+1} e_{2k+1} \dots e_{m(k-k+1)} e_{2e_{k+2}} e_{2k+2} \dots e_{m(k-k+2)}$$
we obtain the initial configuration. The same configuration can also be obtained by performing the null move. Thus, in this case, we have two different global moves that produce the same configuration, so the answer is negative.

If $d = 1$ and $k$ is even, then the global move $e_1 e_2 \dots e_n$ changes each vertex an even number of times, so it has the same effect as the null move. Therefore, in this case, the answer is negative.

Finally, if $d = 1$ and $k$ is odd, we will show that the answer is affirmative. For this, it suffices to show that from the configuration $C = 1, 1, 1, \dots, 1$ we can obtain the configuration $C' = -1, 1, 1, \dots, 1$. Since $(2n, k) = 1$, there exist natural numbers $m$ and $l$ such that $2nm + 1 = kl$. The global move $e_1 e_2 \dots e_{k+1} e_{2k+1} \dots e_{(l-1)(k+1)}$ changes each vertex, except the first one, $2m$ times, and the first vertex is changed $2m + 1$ times, thus transforming $C$ into $C'$.

In conclusion, the answer is affirmative if $(n, k) = 1$ and $k$ is odd, and negative otherwise.