Clearly, if $\cos\phi=\pm1$, we must have $F=\pm id$. Thus we may consider only the case $\cos\phi\not=\pm1$.
A real normal matrix $A$ can be orthogonally block-diagonalized. More specifically, there exists orthogonal matrix $Q$ such that $QAQ^T = (|\lambda_1|R(\theta_1)) \oplus\ldots\oplus(|\lambda_k|R(\theta_k))\oplus \Lambda$, where each $\lambda_i$ is some (perhaps complex) eigenvalue of $A$, $R(\theta)$ means the $2\times 2$ rotation matrix by angle $\theta$, and $\Lambda$ is a real diagonal matrix. See, for instance, Theorem 2.5.8 (p.105) of Horn and Johnson's Matrix Analysis.
Now, if $A$ is orthogonal, we must have $|\lambda_1|=\ldots=|\lambda_k|=1$ and all diagonal entries of $\Lambda$ must be $\pm1$. In other words, for some real orthogonal matrix $Q$, we have $F = QAQ^T = R(\theta_1)\oplus\ldots\oplus R(\theta_k)\oplus I_p\oplus (-I_q)$ for some $k,p,q$. Now, if $\langle Fx, x\rangle = \|Fx\| \|x\|\cos \phi$ for all $x$, similar equalities must also hold blockwise. Therefore, the two blocks $I_p$ and $-I_q$ must be void and $\theta_1,\ldots,\theta_k=\pm\phi$. So your assertion is correct. However, note that $\begin{pmatrix}0&1\\1&0\end{pmatrix}R(-\phi)\begin{pmatrix}0&1\\1&0\end{pmatrix}^T = R(\phi)$. Therefore, we can actually choose an orthonormal basis such the matrix representation of $f$ is $R(\phi)\oplus\ldots\oplus R(\phi)$.