= Solution
Allow an <eigenvector> $v\ne0$ to have complex entries, since real matrices can initially have complex <eigenvalues>. Write $v^*$ for its conjugate transpose. Reality of $A$ gives $A^*=A^T$; symmetry then gives $A^*=A$. Thus $A$ is a <Hermitian matrix>. The scalar $v^*Av$ equals its complex conjugate, because $(v^*Av)^*=v^*A^*v=v^*Av$. If $Av=\lambda v$, then
$$
\lambda=\frac{v^*Av}{v^*v}\in\mathbb R,
$$
since $v^*v>0$. This proves that all <eigenvalues> are real. The equality $A^*=A^T$ is precisely where reality was used; a complex symmetric matrix need not be Hermitian, as $iI$ illustrates.
If $Au=\lambda u$ and $Av=\mu v$, then $u^*Av=\mu u^*v$, but also $u^*Av=(Au)^*v=\overline\lambda u^*v=\lambda u^*v$. Consequently $(\mu-\lambda)u^*v=0$. Distinct <eigenvalues> force $u^*v=0$. For real <eigenvectors> this is ordinary Euclidean orthogonality. These arguments establish <real symmetric spectral orthogonality> without assuming a diagonalization in advance.
A real <orthogonal matrix> satisfies $P^TP=I$, equivalently $P^{-1}=P^T$. If $A$ is symmetric,
$$
(P^{-1}AP)^T=(P^TAP)^T=P^TA^TP=P^TAP=P^{-1}AP,
$$
so orthogonal similarity preserves symmetry. An arbitrary real invertible change of basis need not do so. For example,
$$
A=\begin{pmatrix}1&0\\0&2\end{pmatrix},\qquad P=\begin{pmatrix}1&1\\0&1\end{pmatrix},\qquad P^{-1}AP=\begin{pmatrix}1&-1\\0&2\end{pmatrix}.
$$
The final matrix is not symmetric. \b[Orthogonal similarity preserves real symmetry; general similarity does not.]
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