 Solution

The shift $s:\mathbb{N}\to \mathbb{N}$, $s(n)=n+1$, is injective, so $s\in I(\mathbb{N})$. It is not surjective: the least natural number is missing from its image of a function, whether $\mathbb{N}$ starts at zero or one. For every $t:\mathbb{N}\to \mathbb{N}$, the image of a function of $s\circ t$ is contained in the image of a function of $s$, so $s\circ t\ne {\mathrm{id}}_{\mathbb{N}}$.

The identity function belongs to $I(\mathbb{N})$ and is its identity element under function composition; any group structure using that operation would have this same identity element. Since $s$ has no right inverse element, $I(\mathbb{N})$ is not a group under function composition. It is a monoid, since injectivity is preserved by function composition.