= Solution
A <Lie group> is a finite-dimensional <smooth manifold> with a group structure for which multiplication and inversion are smooth. The group defined here is the <compact symplectic group> $Sp(n)=U(2n)\cap Sp(2n,\mathbb C)$, rather than the full complex symplectic group. Also, the displayed matrix expression requires $B$ to be $2n\times2n$; the printed $n\times n$ size is incompatible with $J$.
Since $J^2=-I$, we have $J^{-1}=-J$. The <matrix exponential> commutes with conjugation, as follows term by term from its absolutely convergent <power series>. Consequently
$$
e^{-JBJ}=e^{JBJ^{-1}}=Je^BJ^{-1}=-Je^BJ.
$$
To construct <logarithm charts for the compact symplectic group>, consider the real <vector space>
$$
\mathfrak k=\{X:X^*=-X,\quad XJ+JX^t=0\}.
$$
Differentiating the defining identities at $I$ gives precisely these conditions. Conversely if $X\in\mathfrak k$, $e^X$ is unitary and
$$
\frac d{dt}(e^{tX}Je^{tX^t})=e^{tX}(XJ+JX^t)e^{tX^t}=0,
$$
so $e^X\in Sp(n)$. These are the infinitesimal conditions of the <compact symplectic Lie algebra>.
Near $I$, the convergent <matrix logarithm> series $\log(I+C)=\sum_{m\ge1}(-1)^{m+1}C^m/m$ is smooth and inverse to the <matrix exponential> near zero. These local inverses respect transpose, conjugate transpose, and conjugation; also $\log(A^{-1})=-\log A$ when both matrices are sufficiently close to $I$. Shrink their neighborhoods accordingly. If $A\in Sp(n)$ there, unitarity gives
$$
(\log A)^*=\log(A^*)=\log(A^{-1})=-\log A.
$$
The symplectic identity is equivalent to $A^t=J^{-1}A^{-1}J$, so, putting $X=\log A$, it gives $X^t=-J^{-1}XJ=JXJ$, equivalently $XJ+JX^t=0$. Thus this local logarithm restricts to a bijection between a neighborhood of $I$ in $Sp(n)$ and an open neighborhood of zero in $\mathfrak k$. Its inverse is the restricted <matrix exponential>. These restrictions are <manifold charts>; left multiplication translates them to every $A_0\in Sp(n)$ via $A\mapsto\log(A_0^{-1}A)$. The chart overlaps are smooth compositions of multiplication, exponential and logarithm. The subspace topology is Hausdorff and second countable, inherited from the finite-dimensional matrix space, so these charts give a <smooth manifold>.
Closure under products and inverses follows from $AJ A^t=J$ and unitarity. Matrix multiplication is polynomial in real and imaginary entries, and inversion on the unitary group is $A\mapsto A^*$, a real linear operation. Their restrictions are smooth in the charts just constructed. Hence $Sp(n)$ is a <Lie group> without needing a closed-subgroup theorem.
Write $X$ in $n\times n$ blocks. The two infinitesimal conditions give
$$
X=\begin{pmatrix}P&Q\\-\overline Q&\overline P\end{pmatrix},\qquad P^*=-P,\quad Q^t=Q.
$$
The <skew-Hermitian matrix> $P$ has $n^2$ real parameters; the complex symmetric matrix $Q$ has $n(n+1)/2$ complex parameters, hence $n(n+1)$ real parameters. Therefore
$$
\boxed{\dim_{\mathbb R}Sp(n)=n(2n+1).}
$$
For $n=1$, any two-by-two matrix satisfies $AJA^t=(\det A)J$, so the group is $SU(2)$. Explicitly,
$$
(a,b)\longmapsto\begin{pmatrix}a&b\\-\overline b&\overline a\end{pmatrix},\qquad |a|^2+|b|^2=1.
$$
The rows are orthonormal and the determinant is one; conversely unitarity and determinant one force this form. This is the <SU(2) as the three-sphere> parametrization. The map and its inverse, extraction of the first row, are smooth. Thus \b[$Sp(1)$ is diffeomorphic to $S^3$.]
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