Degree of a reflection of a sphere

The ambient linear map $R$ is orthogonal, has determinant $-1$, preserves the unit sphere, and satisfies $R^{-1}=R$. Hence its restriction is a smooth diffeomorphism. If $(v_1,\dots ,v_n)$ is a positive tangent basis at $x$, then [F1] makes $(x,v_1,\dots ,v_n)$ positive in ${\mathbb{R}}^{n+1}$. The target outward-normal frame is $(Rx,R{v}_1,\dots ,R{v}_n)=R(x,v_1,\dots ,v_n),$ which has the opposite ambient orientation because $\det R=-1$. Thus the tangent image basis is negative at $Rx$.

Therefore the sphere reflection reverses orientation, and [F2] gives $\mathrm{deg}(R)=-1$. For $n=1$ this is the ordinary reflection of a circle across an axis. Points on the reflecting equator are fixed but still have negative tangent sign, so fixed points do not create a degenerate exception. The disconnected case $S^0$ is excluded by $n\ge 1$; there are no manifold-boundary endpoints, and the pointwise determinant calculation makes no choices.