$E$ an $n$-dimensional vector space. Find all endomorphisms $f$ of $E$ which satisfy $f\circ f = \operatorname{Id}_E$. > Let $E$ be a vector space of dimension $n$. Find all endomorphisms $f$ of $E$ which satisfy $f\circ f = \operatorname{Id}_E$. Is trivial that $f = \operatorname{Id}_E$ is a solution, but I don't know how to see others solutions.

Choose an ordered basis $\mathcal{B}$ for $E$. With respect to this basis, for every endomorphism $f$, there is a matrix $A_f \in \operatorname{M}(n, \mathbb{F})$ such that $[f(e)]_{\mathcal{B}} = A_f[e]_{\mathcal{B}}$, where $[\, \cdot\, ]_{\mathcal{B}}$ means the component vector with respect to $\mathcal{B}$. As $f\circ f = \operatorname{id}_E$, we must have $A_f^2 = I$. Such a matrix is an involutory matrix. In addition, every such matrix defines such an endomorphism, so the number of endomorphisms $f$ that satisfy $f\circ f = \operatorname{id}_E$ is equal to the number of involutive matrices. This number depends on the field $\mathbb{F}$.