Intuitively, this makes sense, but just in $\mathbb{R}^3$.
For example, in $\mathbb{R}^3$, when any $A$ is given, this $A$ determines how much should a point in $\mathbb{R}^3$ be rotated about the origin for each axis. So, if we think of such rotation as a circular movement (which is always happening on some plane,) we can find some normal vector of a circle (plane). And, as $A$ doesn't move any nonzero point on an axis given by the normal vector, we can see that any nonzero point on the $\pm$normal vector will satisfy $Av=\pm v$.
However, I am having trouble showing why this works for every odd $n$. Also, I wonder how I should use my approach to write a mathematical proof.
Maybe I should bring the concept of the linear hyperplane $H$ to use my approach. As $A$ gives us an orthonormal basis of $\mathbb{R}^n$, I guess we could eliminate one of the elements in the basis to create $H$, which is a similar process that I used.
Moreover, I wonder we can say this is true only when $n$ is odd because we get a counterexample when $n=2$, right?
Any help would be really appreciated.
Heh. Does it, though? How would you describe what all the elements of $O(3)$ are? The result is obvious given that description, but you have to do at least a little work to get the description.
Anyway, you can show the result with algebra. The characteristic polynomial has odd degree, so it has a real root $\lambda$, in which case $A-\lambda I$ has nontrivial kernel in $\mathbb{R}^n$, so there is an eigenvector $v$ with $Av=\lambda v$. Since $A$ is an isometry, you can take the norm of both sides to get $|\lambda|=1~\Rightarrow~\lambda=\pm1$.
More generally, you can massage the spectral theorem to tell you every $A\in O(V)$ (where $V$ is an inner product space) is, in some basis, equivalent to a block-diagonal matrix, where the blocks are $2\times2$ rotation matrices with possibly one $1\times1$ block $[\pm1]$ in the corner (depending on $\det A=\pm1$).