If there is an orthogonal basis for $\Bbb R^n$ consisting of eigenvectors of $H$, then $H^T=H$?

Question

Question: For any $n\times n$ real matrix $H$, if there is an orthogonal basis for $\Bbb R^n$ consisting of eigenvectors of $H$, then $H^T=H$?

Does this mean that $H$ is unitarily diagonalizable? If so, aren't orthogonal or skew symmetric matrices unitarily diagonalizable as well, so why is $H^T=H$(symmetric) true?