If $T : V \to V$ is a linear map and $H \le V$ a subspace, such that $T_H = \operatorname{id}_H$, i.e $T(v) = v$ for all $v \in H$ and also that $T$ induces the identiy transformation on $V / H$, i.e. $\overline{T}(v + H) = v + H$ or equivalently $Tv - v \in H$ for all $v \in V$.
Then do we have $\det T = 1$?
If $e_1,\dots,e_k$ is a basis of $H$ and $\hat f_{k+1},\dots,\hat f_n$ a basis of $V/H$, then $e_1,\dots,e_k,f_{k+1},\dots,f_n$ is a basis of $V$. From $T(\hat f_j)=\hat f_j$ it follows $T(f_j)-f_j\in H$, so we can write $T(f_j)=f_j+$ a linear combination of $e_1,\dots,e_k$. Thus the matrix of $T$ in this base has the form $\left(\begin{matrix}I_k&0\\*&I_{n-k}\end{matrix}\right)$ and therefore $\det T=1$