Solution
= Solution
{c}
An <injective> <function> from a <finite set> to itself is <surjective>: its <image of a function> contains as many distinct elements as the entire <set>, so it cannot omit one. Thus every element of $I(X)$ is a <bijection>, and conversely every <bijection> is <injective>. Hence
$$
\boxed{I(X)=\operatorname{Sym}(X)\quad\text{for finite }X.}
$$
It is a <group> under <function composition> by <5/a/i>.