 Solution

A nonidentity transvection is a linear automorphism $\tau =I+N$ with $\mathrm{rank}N=1$ and $N^2=0$. Equivalently, it fixes a hyperplane pointwise and its displacement has image in a line lying in that hyperplane. The notation $T\setminus \{1\}$ allows the identity as a degenerate transvection; use that convention here, with $N=0$ also allowed.

For a nonidentity transvection, choose $d\ne 0$ spanning the image of $N$. There is a unique nonzero linear functional $f$ such that $N(v)=f(v)d$. Since $N^2(v)=f(v)f(d)d$, we have $f(d)=0$, so $d\in \ker f$. Conversely, this condition makes $(df{)}^2=0$ and gives inverse $I-df$. The identity is represented by $f=0$ and any $d\ne 0$. ThusWhen $f(d)=f^{\prime }(d)=0$, the two displacement maps have zero product, givingFor clarity use left actions of linear maps. Applying $g^{-1}$, then ${\tau }_{f,d}$, then $g$ givesWith the alternative conjugation notation ${\tau }^g=g^{-1}\tau g$, the answer is ${\tau }_{f,d}^g={\tau }_{f\circ g,g^{-1}(d)}$. In either convention the transformed direction lies in the kernel of the transformed functional.