I will denote the $n-$sphere of radius $1$ centered at the origin as $\mathbb{S}^n$, so that $$ \mathbb{S}^n = \{ x \in \mathbb{R}^{n+1}\ : \ \|x\| = 1\}. $$ I am stuck on the following problem...I'm not even sure where to begin. I'm not sure I understand the definition of $g$, and how we can extrapolate information about $g$ from that definition. Also, what would $\mathbb{S}_a^{n-1}$ look like as defined below?
We will treat $\mathbb{R}^n$ as sitting inside $\mathbb{R}^{n+1}$ as the set of points whose last coordinate is $0$. We then have $\mathbb{S}^{n-1}$ sitting inside $\mathbb{S}^n$ as the equatorial sub-sphere. Suppose $a \in \mathbb{S}^n$ and let $\mathbb{S}_a^{n-1}:= a^{\perp} \cap \mathbb{S}^n.$ Suppose that $g \in O(n+1)$ and $g(a) = e_{n+1}$. Suppose $f \in O(n+1)$ and $f(a)=a$, so $f$ maps $\mathbb{S}_a^{n-1}$ isometrically to itself.
$(a)$ Explain why $g$ maps $\mathbb{S}_a^{n-1}$ isometrically to $\mathbb{S}^{n-1}$.
$(b)$ Explain why $gfg^{-1} \in O(n)$ and is therefore an isometry of $\mathbb{S}^{n-1}$.
$(a)$ Let $x \in a^\perp$. Then $\langle x,a \rangle = 0$. Since $g$ is orthogonal, we also have $\langle g(x),g(a) \rangle = 0 \iff \langle g(x),e_{n+1} \rangle = 0$. Thus $g(x) \in e_{n+1}^\perp = \mathbb{R}^n$.
Now let $x \in \mathbb{S}_a^{n-1}$. Then the above condition holds, and also $\|g(x)\| = \|x\| = 1$. Hence $g(x) \in \mathbb{S}^{n-1}$. Hence $g$ maps $\mathbb{S}_a^{n-1}$ isometrically to $\mathbb{S}^{n-1}$.
$(b)$ Since $g$ is orthogonal, $g^{-1}$ is also orthogonal. And since $gfg^{-1}$ is a composition of orthogonal matrices, it is also orthogonal. Note the following: \begin{align*} g^{-1}(e_{n+1}) & = a \\ fg^{-1}(e_{n+1}) & = f(a) = a \\ gfg^{-1}(e_{n+1}) & = g(a) = e_{n+1}. \end{align*} Since $gfg^{-1}(e_{n+1}) = e_{n+1}$, we conclude that if $x \in S^{n-1} = e_{n+1}^\perp \cap \mathbb{S}^n$, then $0 = \langle x, e_{n+1} \rangle$ $= \langle f(x), f(e_{n+1}) \rangle = \langle f(x), e_{n+1} \rangle$. Hence $x \in e_{n+1}^\perp$. But also $1 = \|x\| = \|f(x)\|$, so $x \in \mathbb{S}^n$. Thus we conclude that $x \in e_{n+1}^\perp \cap \mathbb{S}^n = S^{n-1}$, that is, $gfg^{-1}$ is an isometry of $\mathbb{S}^{n-1}$.