An affirmative answer is claimed on page 3 here (unless I misunderstand),
Lemma: If two symmetric matrices $A$ and $B$ are similar, then they are orthogonally similar.
but I was unable to find a proof or citation.
To understand this claim I considered the simple case of
$$ A = \left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & i \\ 1 & i & 0 \\ \end{array} \right), \qquad B = \left( \begin{array}{ccc} i \sqrt{2} & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & -i \sqrt{2} \\ \end{array} \right), $$ which (by virtue of their Jordan normal form) are both similar to the matrix $$ J = \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{array} \right). $$ The aforementioned Lemma then claims that $A$ and $B$ are related by a similarity transformation $A = P B P^{-1}$, such that $P$ is equal to its inverse transpose: $P = P^{-T}$. Following pointers from @chhro I see that $$ P = \frac{1}{4}\left( \begin{array}{ccc} -i & -3 \sqrt{2} & i \\ -3 & i \sqrt{2} & 3 \\ -2 \sqrt{2} & 0 & -2 \sqrt{2} \\ \end{array} \right) $$ is a solution. However I would be grateful for reference for the general claim.
As @chhro mentions the answer to the question is yes, and the proof follows from Theorem 20 of Horn and Merino. However, a more direct proof is also provided in the later work of Bukovšek and Omladic:
We assume $A = A^T$, $B = B^T$, $A = P B P^{-1}$ and wish to construct complex orthogonal matrix $O$ ($OO^T = I$) such that $A = O B O^T$.
From the premise we have $A P = P B$, by transposition we have $P^T A = B P^T$, and multiplication by $P$ we have $P P^T A = P B P^T = A P P^T$ hence $A$ and $PP^T$ commute.
Let $R$ be the square root of the complex symmetric invertible matrix $PP^T$: $R^2 = PP^T$, $R$ is symmetric, invertible, and commutes with $A$.
Let $O = R^{-1}P$. $O$ is complex orthogonal $OO^T = R^{-1}P P^T R^{-1} = R^{-1}R^2 R^{-1} = I$. Moreover, $O$ relates $A$ and $B$ in the manner desired: $A O = A R^{-1} P = R^{-1} A P = R^{-1} P B = O B$.