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I can prove that any elemento $A\in SO(3)$ must be the conjugation of: $$A'=\left(\begin{array}{ccc} \cos\theta &-\sin\theta &0\\ \sin\theta &\cos\theta &0\\ 0&0&1 \end{array}\right)$$ (that is, there exists $P\in O(3)$ s.t. $P^t AP=A'$)

My question is: $P$ must be the identity matrix? (and so $A=A'$)?

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    Your question is strange: evidently $P$ has not to the identity: it suffices to take $P=\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 &\cos\alpha &-\sin\alpha \\ 0 & \sin\alpha &\cos\alpha \end{array}\right)$ – 2017-02-04

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No, $P$ need not be the identity matrix. In general, an element $A\in SO(3)$ has the form

$$ \left( \begin{array}{ccc} \cos\theta \cos\psi & -\cos\phi \sin\psi + \sin\phi \sin\theta \cos\psi & \sin\phi \sin\psi + \cos\phi \sin\theta \cos\psi \\\ \cos\theta \sin\psi & \cos\phi \cos\psi + \sin\phi \sin\theta \sin\psi & -\sin\phi \cos\psi + \cos\phi \sin\theta \sin\psi \\\ -\sin\theta & \sin\phi \cos\theta & \cos\phi \cos\theta \end{array} \right) $$