Let $c$ be the cycle $i \mapsto i - 1 \bmod n$. The pairs obeying your condition are of the form $$((\overbrace{x,x,x,\ldots,x}^{n-k},\overbrace{x+1,x+1,\ldots,x+1}^{k}), c^k).$$ Under the group multiplication that you write out, they are isomorphic to $\mathbb{Z}$, with generator $((0,0,0,\ldots,0,1), c)$.

It is easy to check that $((0,0,0,\ldots,0,1), c)$ obeys the given conditions, and hence its powers do. We now must check that any $(v, \sigma)$ obeying these condition is a power of $((0,0,0,\ldots,0,1), c)$.

Using your group operation, we can replace $\sigma$ by $c^k \sigma$ for any $k$, and therefore we can assume that $\sigma(1) = 1$. For any $j$, we then have $\sigma(1) < \sigma(j)$, so $v(1) = v(j)$, so $v$ is of the form $(x,x,\ldots, x)$, and we are done.

Let $c$ be the cycle $i \mapsto i+ 1 \bmod n$. The pairs obeying your condition are of the form $$((\overbrace{x+1,x+1,x+1,\ldots,x+1}^{k},\overbrace{x,x,\ldots,x}^{n-k}), c^k).$$ Under the group multiplication that you write out, they are isomorphic to $\mathbb{Z}$, with generator $((0,0,0,\ldots,0,1), c)$.

It is easy to check that $((0,0,0,\ldots,0,1), c)$ obeys the given conditions, and hence its powers do. We now must check that any $(v, \sigma)$ obeying these condition is a power of $((0,0,0,\ldots,0,1), c)$.

Using your group operation, we can replace $\sigma$ by $c^k \sigma$ for any $k$, and therefore we can assume that $\sigma(1) = 1$. For any $j$, we then have $\sigma(1) < \sigma(j)$, so $v(1) = v(j)$, so $v$ is of the form $(x,x,\ldots, x)$, and we are done.

The extended Weyl group of $GL_n$ is often described as the group of bijections $f:\mathbb{Z} \to \mathbb{Z}$ obeying the condition $f(i+n)=f(i)+n$. Then $\sigma$ is the reduction modulo $n$, and I'm not going to try to get the indices right to figure out what $v$ is. The length $0$ elements are the permutations of the form $i \mapsto i+k$.