Does $A$ is similar to a symmetric matrix $\implies $ $A$ is symmetric? Let $A\in \mathbb R^{n\times n}$. Is it true that 

$A$ similar to a symmetric matrix $\implies $ $A$ symmetric ?

Let $B$ symmetric s.t. $A=PBP^{-1}$. Then $$A^T=(P^{-1})^TB^T P^T=(P^{-1})^T BP^T.$$
For me there is no reason that $P$ is orthogonal, so I would say it's false a priori. But in the same time, this theorem should be true since   operator is self adjoint $\iff$ it's diagonalizable. I also know that matrices in any basis of Self Adjoint operator are symmetric. But if A is similar to a symmetric matrix, then it's diagonalizable and thus self adjoint, and thus, it should be symmetric in any basis... this is wrong ? If yes, why ?
 A: Take any $A$ which is diagonizable but not symmetric. So $A = TDT^{-1}$ for a diagonal matrix $D$ and invertible $T$. For any orthogonal matrix $O$ $A$ is similar to $ODO^\intercal$, which is symmetric, but $A$ is not.
A: Try a $P$ that is not orthogonal.
Let me try $P=\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}$, $P^{-1}=\begin{bmatrix} \frac12 & 0 \\ 0 & 1\end{bmatrix}$
Let $B=\begin{bmatrix} 1 & 1 \\ 1 & 1\end{bmatrix}$
then $$A=PBP^{-1}=\begin{bmatrix} 2 & 0 \\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 1\end{bmatrix}\begin{bmatrix} \frac12 & 0 \\ 0 & 1\end{bmatrix}=\begin{bmatrix} 2 & 0 \\ 0 & 1\end{bmatrix}\begin{bmatrix} \frac12  & 1\\ \frac12 & 1\end{bmatrix}=\begin{bmatrix} 1 & 2\\ \frac12  & 1
\end{bmatrix}$$
which is not symmetrical.
A: Since there is already a counterexample in the other answer, I will focus on this:

But in the same time, this theorem should be true since symmetric matrices looks very important, and such a result with this type of matrix should hold...

It is true that symmetric matrices (with real entries) are important because they correspond to self-adjoint linear operators. Such operators are always* diagonalizable in an orthonormal basis and have real eigenvalues. Moreover, those attributes characterize self-adjoint operators, i.e. any operator diagonalizable in an orthonormal basis with real eigenvalues is self-adjoint. 
So basically a symmetric matrix is important because of its properties which relate to the inner product. But similarity of matrices does not preserve the inner product structure (only conjugation with unitary matrices does), hence it also does not necessarily preserve symmetric matrices. A similar example in topology is that a space homeomorphic to a complete metric space need not be complete.

*In this answer I am only referring to finite-dimensional unitary spaces.
A: 
An operator over $\mathbb{R}^n$ is self-adjoint if and only if it is diagonalizable with an orthogonal matrix.

You are missing the clause shown in italics.
For instance, the matrix
\begin{bmatrix}
1 & 1 \\
0 & 2
\end{bmatrix}
is diagonalizable, having two distinct eigenvalues ($1$ and $2$). Hence the matrix is similar to a diagonal, hence symmetric, matrix. However, this matrix is not orthogonally similar to a diagonal matrix, because its basic eigenvectors are not orthogonal to each other.

If $A$ is diagonalizable with an orthogonal matrix, that is
$$
A=PDP^T
$$
with $D$ diagonal and $P^T=P^{-1}$, then
$$
A^T=(PDP^T)^T=PD^TP^T=PDP^T=A
$$
The converse ($A$ symmetric is diagonalizable with an orthogonal matrix) is a deeper theorem, a consequence of the more general spectral theorem.
