I know that the isometries of $\mathbb{S}^n\subseteq\mathbb{E}^{n+1}$ are $O(n+1)$. Intuitively, I think that $SO(n+1)$ should be the orientation-preserving ones. But I'd like to prove this rigourosly.
I know that a map is orientation preserving iff its differential at any point is orientation preserving. Clearly if $A\in O(n+1)$ is an isometry, then $d_pA(v)=Av$ but $d_pA$ is NOT an endomorphism (it has different domain and codomain) so I cannot simply say that it's orientation preserving iff $\det(A)>0$. Right?
How do I solve this?
Let $S^n\subset \Bbb R^{n+1}$ be the unit sphere. Notice that if $p\in S^n$, then $T_pS^n = \{p\}^{\perp} \subset \Bbb R^{n+1}$, so that $p$ is the outward pointing unit normal vector to $S^n$ at $p$.
Consider the volume form $\omega$ on $S^n$ defined by $$ \forall p \in S^n, \forall v_1,\ldots,v_n \in T_pS^n,\quad \omega_p(v_1,\ldots,v_n) = \det(p,v_1,\ldots,v_n). $$ Let $A$ be an isometry of the sphere, which is the restriction of a linear isometry in $O(n+1)$, that we still denote $A$. As you noticed, it follows from the linearity of $A$ that $d_pA = A$ seen as a linear map $T_pS^n=\{p\}^{\perp} \to T_{Ap}S^n = \{Ap\}^{\perp}$. In other words, $$ \forall p \in S^n,\forall v \in T_pS^n,\quad d_pA(v) = Av \in T_{Ap}S^n. $$ (One can directly check that $Av\in T_{Ap}S^n$, since $\langle Ap,Av\rangle = \langle p,v\rangle = 0$, and $Av \in \{Ap\}^{\perp} = T_{Ap}S^n$). Hence, one has \begin{align} \forall p\in S^n, \forall v_1,\ldots,v_n \in T_pS^n,\quad (A^*\omega)_p(v_1,\ldots,v_n) &= \omega_{Ap}(d_pA(v_1),\ldots,d_pA(v_n)) \\ &= \omega_{Ap}(Av_1,\ldots,Av_n)\\ &= \det(Ap,Av_1,\ldots,Av_n)\\ &= (\det A) \det (p,v_1,\ldots,v_n)\\ &= (\det A)\, \omega_p(v_1,\ldots,v_n), \end{align} where the fourth equality is a well-known property of the determinant. Finally, $A^*\omega = (\det A) \, \omega$, and it follows that $A$ preserves the orientation induced by $\omega$ if and only if $\det A > 0$, that is, if and only if $A \in SO(n+1)$.