When $A \in SO(3)$, $A$ is always a rotation. I have official proof of this problem, but I am having trouble understanding some parts of it. Thus, I would like to share the parts and would like to get checked if my understanding is correct.
Before beginning, I'd like to make sure all the notations are clear.

*

*The space is $\mathbb{R}^3$.

*$SO(3)$ is the special orthogonal group of $\mathbb{R}^3$. Thus, any $A \in SO(3)$ must have $1$ as its determinant.

*The solution will use the fact "For odd $n \in \mathbb{N}$, There is $v \in \mathbb{R}^n$ s.t $v \neq 0$ and $Av=\pm v$ for every $A \in O(n)$ "

Solution:

As $3$ is an odd number, for any $A \in SO(3)$, there is a nonzero $v \in \mathbb{R}^3$ s.t $Av= \pm v$.
This gives us the information about the eigenvalue of $A$, which is $\lambda = \pm 1$.
Let $l$ be a line parallel to $v$ going through $0$, then let the perpendicular hyperplane be $H$.
Then, $A(H)=H$, and also $A$ would be restricted to $\underline{A}$ which is a linear isometry of $H$.

*

*Case 1: $\lambda =1$.

As $\lambda =1$, $\underline{A}$ is a direct isometry on $H$.
As $H$ is a plane, $\underline{A}$ is clearly a rotation because it can't be a translation.
Then, $A$ is a rotation about $l$.

*

*Case 2: $\lambda = -1$.

For a similar reason, $\underline{A}$ is an opposite isometry of $H$.
Then, $\underline{A}$ is a reflection in some line $l'$ in $H$ going through $0$.
Then, $A$ is rotation about $l'$ by $\pi$
Indeed, when $\lambda =1$, $A$ is conjugate to $\begin{bmatrix}
 \cos \alpha& - \sin \alpha & 0 \\
  \sin \alpha& \cos \alpha & 0 \\
0 & 0 & 1 \\ 
\end{bmatrix}$
And, when $\lambda = -1$, $A$ is conjugate to $\begin{bmatrix}
 1& 0 & 0 \\
  0& -1 & 0 \\
0 & 0 & -1 \\ 
\end{bmatrix}$.

Questions:

*

*My textbook definition of direct and opposite isometries: For $T \in I(\mathbb{R}^n)$ s.t $T(x)=Ax+a$ with $A \in O(n)$ and $a \in \mathbb{R}^n$, we say $T$ is direct if $\det (A)=1$ and opposite if $\det(A) =-1$.

But, I wonder how the eigenvalue of $A$ in the solution could imply that $\underline{A}$ is direct or opposite. I guess direct or opposite also implies that the $A \in I(\mathbb{R}^3)$ preserves orientation or not, so as $\underline{A}$ is just a restricted version of $A$, $\underline{A}$ must preserve orientation if $A$ does, otherwise it doesn't. Am I correct?
Moreover, if $A \in O(3)$ is opposite, then $\lambda =1$ implies $\underline{A}$ is opposite. And, $\lambda =-1$ implies that $\underline{A}$ is direct. But, in this case, why do get a different result from the solution here?

*

*Conjugate

Actually, this is the part which I was really curious about.
Does $A$ being conjugate to the matrix $\begin{bmatrix}
 \cos \alpha& - \sin \alpha & 0 \\
  \sin \alpha& \cos \alpha & 0 \\
0 & 0 & 1 \\ 
\end{bmatrix}$ mean $A$ doesn't move any point on the axis $l$, but rotates any other point on $H$ for $\alpha$?
If so, $$A = B \begin{bmatrix}
 \cos \alpha& - \sin \alpha & 0 \\
  \sin \alpha& \cos \alpha & 0 \\
0 & 0 & 1 \\ 
\end{bmatrix} B^{-1}$$, and $B$ must be some rotation in $\mathbb{R}^3$ that makes axis $z$ the line $l$. Is it correct way of understanding this solution?
So, in general, does the conjugate matrix always tell us the properties of $A \in I(\mathbb{R}^n)$ in some hyperplane $H$ of $\mathbb{R}^n$?
 A: 
I guess direct or opposite also implies that the $A \in I(\mathbb{R}^3)$ preserves orientation or not, so as $\underline{A}$ is just a restricted version of $A$, $\underline{A}$ must preserve orientation if $A$ does, otherwise it doesn't. Am I correct?

Not quite. $A$ preserve orientation of $\mathbb R^3$ does not imply that $A|_H$ preserve the orientation of $H$. Since $\det A = \lambda \det (A|_H) = 1$, we have $\det(A|_H)=\frac{1}{\lambda}$ and $A|_H$ preserves orientation iff $\lambda =1$.

Does $A$ being conjugate to the matrix $\begin{bmatrix}
 \cos \alpha& - \sin \alpha & 0 \\
  \sin \alpha& \cos \alpha & 0 \\
0 & 0 & 1 \\ 
\end{bmatrix}$ mean $A$ doesn't move any point on the axis $l$, but rotates any other point on $H$ for $\alpha$?

Yes.

If so, $$A = B \begin{bmatrix}
 \cos \alpha& - \sin \alpha & 0 \\
  \sin \alpha& \cos \alpha & 0 \\
0 & 0 & 1 \\ 
\end{bmatrix} B^{-1}$$, and $B$ must be some rotation in $\mathbb{R}^3$ that makes axis $z$ the line $l$. Is it correct way of understanding this solution?

Yes.
I guess your confusion with the solution is when say $A=\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ (i.e. $\alpha=\pi$), the proof doesn't necessarily produce $H=\{(x,y,0)|x,y\in\mathbb R\}$, but possibly $H=\{(0, y, z)|y, z\in\mathbb R\}$ in which case the action of $A_H$ doesn't preserve the orientation.
Personally, I prefer a more algebraic proof: as $A^tA=AA^t=I$, $A$ is unitary, hence diagonalizable over $\mathbb C$. Note that any eigenvalue of $A$ must satisfy $|\lambda|=1$, and $A$ must have at least $1$ real eigenvalue, and the eigenvalues of $A$ are closed under conjugation, combined with $\det A=1$, we must have the eigenvalues of $A$ must be $1, e^{i\alpha}, e^{-i\alpha}$.
