= Solution
{c}
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 $s\circ t\ne\operatorname{id}_{\mathbb N}$.
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>, \b[$I(\mathbb N)$ is not a <group> under <function composition>]. It is a <monoid>, since <injectivity> is preserved by <function composition>.
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