 Solution

Let $v=T(0)$ and $B(x)=T(x)-v$. The Euclidean isometry $B$ fixes the origin and preserves norms. The polarization identity therefore gives $⟨Bx,By⟩=⟨x,y⟩$. In particular, $B{e}_1,B{e}_2$ are an orthonormal basis. The coefficients of $Bx$ in this basis are $⟨Bx,B{e}_j⟩=x_j$, so $Bx=x_{1}B{e}_1+x_{2}B{e}_2$. Thus $B$ is a linear map represented by an orthogonal matrix, and $T(x)=Bx+v$. Conversely, every such map preserves distances because $\Vert B(x-y)\Vert =\Vert x-y\Vert$.

When $\det B=-1$, the real eigenvalues of the two-dimensional orthogonal matrix are $1,-1$. Choose corresponding unit eigenvectors $e,n$, and write $v=ae+bn$. In these coordinates $T(s,t)=(s+a,-t+b)$. This is reflection in the line $t=b/2$, followed by translation $ae$ along that line. It is a Euclidean reflection if $a=0$, and a glide reflection if $a\ne 0$.