See the answer here.
Over the reals the intersection of the orthogonal and symplectic groups in even dimension is isomorphic to the unitary group in the half dimension:$$ U(n) = O(2n, \mathbf{R}) \cap Sp(2n, \mathbf{R}).$$
To me, this is not so obvious. How do we see this without just saying "oh, that's trivial, cite the "2-out-of-3 property"?
The result follows from carefully unwinding the identifications involved and the definitions of the various groups. Let $V = \mathbb{R}^{2n}$ and let us denote elements in $V$ by $(x_1, \ldots, x_n, y_1, \ldots, y_n)$. Define a map $\Phi \colon V \rightarrow \mathbb{C}^n$ by
$$ \Phi(x_1, \ldots, x_n, y_1, \ldots, y_n) = (x_1 + \sqrt{-1}y_1, \ldots, x_n + \sqrt{-1}y_n). $$
The map $\Phi$ is an $\mathbb{R}$-linear isomorphism that allows us to identify $V$ with $\mathbb{C}^n$, pullback all the structure of $\mathbb{C}^n$ to $V$ and think about $M_n(\mathbb{C})$ as sitting inside $M_{2n}(\mathbb{R})$. Let us make this explicit. Denote by $J$ the matrix
$$ J = \left( \begin{matrix} 0_{n \times n} & -I_{n \times n} \\ I_{n \times n} & 0_{n \times n} \end{matrix} \right) \in M_{2n}(\mathbb{R}). $$
Using the identification of $\Phi$, the linear map $T_J$ (the map represented in the standard basis by $J$) is the pullback of the natural complex structure on $\mathbb{C}^n$. A real $(2n)\times(2n)$ matrix $E \in M_{2n}(\mathbb{R})$ (whose elements correspond to a $\mathbb{R}$-linear map of $V$) belongs to $M_{n}(\mathbb{C})$ (whose elements correspond to a $\mathbb{C}$-linear map of $\mathbb{C}^n$) if and only if $EJ = JE$ (that is, $T_E$ commutes with $T_J$ and so the map $\Phi \circ T_E \circ \Phi^{-1}$ is complex linear).
If we write $E$ as a $2 \times 2$ block matrix with blocks of size $n$, we have that $E \in M_n(\mathbb{C})$ if and only if $E$ has the form
$$ E = \left( \begin{matrix} A & -B \\ B & A \end{matrix} \right) $$
with $A, B \in M_n(\mathbb{R})$ and the matrix $E$ is then identified with the matrix $A + \sqrt{-1}B \in M_n(\mathbb{C})$. The matrix $A + \sqrt{-1}B$ is unitary if and only if
$$(A + \sqrt{-1}B)^{*} (A + \sqrt{-1}B) = (A^t - \sqrt{-1}B^t)(A + \sqrt{-1}B) = (A^tA + B^t B) + \sqrt{-1}(A^tB - B^tA) = I + \sqrt{-1} \cdot 0. $$
That is, if and only if $A^tA + B^t B = I$ and $A^t B - B^t A = 0$.
Now, let $E \in O(2n,\mathbb{R}) \cap \mathrm{Sp}(2n, \mathbb{R})$ and write
$$ E = \left( \begin{matrix} A & C \\ B & D \end{matrix} \right) $$
with $A,B,C,D \in M_n(\mathbb{R})$. Then $E^tE = I$ (because $E \in O(2n, \mathbb{R})$) and $E^tJE = J$ (because $E \in \mathrm{Sp}(2n, \mathbb{R})$). Multiplying the last equation by $E$ from the left, we get $EE^tJE = JE = EJ$ and so $E \in M_{n}(\mathbb{C})$. Thus, $C = -B$ and $D = A$. But then, writing the orthogonality condition we see that
$$ E^t E = \left( \begin{matrix} A^t & B^t \\ -B^t & A^t \end{matrix} \right) \left( \begin{matrix} A & -B \\ B & A \end{matrix} \right) = \left( \begin{matrix} A^tA + B^tB & -A^tB + B^tA \\ -B^tA + A^tB & B^tB + A^tA \end{matrix} \right) = \left( \begin{matrix} I & 0 \\ 0 & I \end{matrix} \right) $$
which shows that $A + iB$ is unitary and thus $E \in U(n)$. I'll leave you the details for the other direction.