$T:V\rightarrow V$ is over $\mathbb{R}$ , it's matrix is $A$, $A=PDP^*$. Is it true that $A$, $D$, and $P$ are in $M_{n \times n}(\mathbb{R})$

$T:V\rightarrow V$ is over $\mathbb{R}$ and $V$ of finite dimension $n$, and I know that it is orthogonally diagonalizable.

The Matrix that represents it - call it $A$ ,in orthonormal basis is orthogonally diagonalizable as well. So $A=PDP^*$ where $P^*$ is a unitary matrix, and $D$ a diagonal matrix.

Is it true that $A$, $D$, and $P$ are in $M_{n \times n}(\mathbb{R})$ (real matrices) because the transformation is over $\mathbb{R}$? I want to conclude that $P^*$ = $P^t$.

Thanks a lot!!

Let $A=\left(\begin{matrix}0&1\\ -1&0\end{matrix}\right)$.
The $A$ is anti-symmetric, and therefore it is orthogonally diagonalizable, and its eigenvalues are $\pm i$. Therefore, $D$ contains purely imaginary elements in the diagonal. Clearly, $P$ is not a real matrix either. In fact $$P=\left(\begin{matrix} 1& i\\ i&1\end{matrix}\right)$$
Nevertheless it is true is $A$ is symmetric (real symmetric). In such case, the eigenvalues are real (and so is $D$) and a diagonalization is achieved via a real unitary matrix $P$.