Show $f(W) \subset W \Rightarrow f(W^\perp) \subset W^\perp$ and $f(\langle v \rangle ) \subset \langle v \rangle \Rightarrow$ eigenvector Let $f \in \textrm{O}(n, \mathbb R)$. (O is the orthogonal group) Show:
i) If $f(W) \subset W$ for a subspace $W \subset \mathbb R^n$, then $f(W^\perp) \subset W^\perp$.
ii) If $f(\langle v \rangle ) \subset \langle v \rangle$ for a vector $0 \neq v \in \mathbb R$, then $v$ is an eigenvector of $f$ with value $1$ or $-1$.
 A: (i) Let $x\in W^\perp$ and $y\in W$ then
$$\langle {f(x)},f(y)\rangle=\langle x,y\rangle=0$$
and since $f(y)\in W$ hence we conclude that $f(x)\in W^\perp$.
(ii) By  the hypothesis there's $\lambda$ such that $f(v)=\lambda v$ and then
\begin{align}\langle f(v),f(v)\rangle&=\langle \lambda v,\lambda v\rangle=|\lambda|^2||v||^2\\&=\langle v,v\rangle=||v||^2\end{align}
hence since $||v||\ne0$ then $|\lambda|^2=1$. Conclude.
A: Note that since $f \in O(n, \Bbb R)$, that is, $f$ is an orthogonal linear map, we have $f^Tf = ff^T = I$, where $I$ is the $n \times n$ identity map.  Suppose $\dim W = m$.  Then, since $f$, being orthogonal, is nonsingular, we have $\dim f(W) = \dim W = m$ as well; this shows that in fact $f(W) = W$, so that for any $y \in W$ there is a unique $z \in W$ with $y = f(z)$.  Now let $x \in W^\bot$; then $\langle x, y \rangle = 0$ for all $y \in W$.  Consider $\langle f(x), y \rangle$; we have, since $y = f(z)$ with $z \in W$, 
$\langle f(x), y \rangle = \langle f(x), f(z) \rangle = \langle x, f^Tfz \rangle = \langle x, z \rangle = 0, \tag{1}$
since $f^Tf = I$.  Alternatively, one can argue directly that 
$\langle f(x), f(z) \rangle = \langle x, z \rangle = 0 \tag{2}$
since $f \in O(n, \Bbb R)$ is an isometry, that is, it preserves the inner product $\langle \cdot, \cdot \rangle$.  Either way, we have that $\langle f(x), y \rangle = 0$ for all $y \in W$; thus $x \in W^\bot$, and thus $f(W^\bot) \subseteq W$.  In fact, we must have $f(W^\bot) = W^\bot$ by the same dimensionality argument used to show $f(W) = W$.
If $f(\langle v \rangle) \subset \langle v \rangle$, we note in passing that once again we may conclude $f(\langle v \rangle) = \langle v \rangle$ by the same logic which showed $f(W) = W$ etc., but that in any event $f(v) = \mu v$ for some $\mu \in \Bbb R$; thus
$\langle v, v \rangle = \langle f(v), f(v) \rangle = \langle \mu v, \mu v \rangle = \mu^2 \langle v, v \rangle; \tag{3}$
now $v \ne 0$ forces $\mu = \pm 1$ and we are done.  QED.
Note:  $f \in O(n, \Bbb R)$ means, by definition,  $\langle f(x), f(y) \rangle = \langle x, y \rangle$ and/or $f^Tf = ff^T = I$, since $\langle f(x), f(y) \rangle = \langle x, f^Tfy \rangle$ for all $x, y \in \Bbb R^n$, whence $f^Tf = I$ etc.  Whether one uses the language $f^T$ (the transpose or "adjoint" of $f$) or not, the result is the same since $f$ preserves $\langle \cdot, \cdot \rangle$.  End of Note.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
