If a matrix is orthogonal, then it is normal. Then why isn't an orthogonal matrix orthogonally diagonizable

If P is orthogonal then (P^T) = (P^-1) =>
(P^T)(P) = I = (P)(P^T) => P is Normal => P is orthogonally diagonizable => P is symetric P = (P^T)

But an orthogonal matrix is not always symetric. What am I doing wrong?

Awms example doesn't address the issue here because his matrix isn't orthogonal. Moreover, it isn't orthogonally diagonalizable, it is unitarily diagonalizable.

The problem is here "P is normal ==> P is orthogonally diagonalizable". The correct statement is

P is normal ==> P is UNITARILY diagonalizable.

If P has complex eigenvalues, then note that P unitarily diagonalizable ensures the existence of unitary matrix U such that

P = U*DU.

Then

P* = P^T = U*D*U.

But in this case, D* need not equal D!

If P does happen to be symmetric, we know that the eigenvalues must be pure real, but we are not assuming P is symmetric.

Oh, your addition is correct. A real matrix P is symmetric if and only if it is orthogonally diagonalizable.

The problem is this step:
"P is orthogonally diagonizable => P is symetric P = (P^T)"

This is not true.

For instance, the following matrix is normal, and hence orthogonally diagonalizable, but isn't symmetric:
(1 1 0)
(0 1 1)
(1 0 1)