$U(n)$ is a subgroup of $Sp(n)$ fixed by an involution I read in a paper this  passage 
"We can view $U(n)$ as a subgroup of $Sp(n)$ which is fixed by an involution $\theta$ ..."
My question is what is this involution, please  ?
 A: EDIT: Jason DeVito's answer made me realize that the question is probably more about the compact symplectic group $Sp(n)$ than about the noncompact symplectic group $Sp(2n)$ I consider in my answer below (by "déformation professionnelle" I guess). I leave my answer for reference in any case and I also added another solution inspired by Jason DeVito's answer.
The existence of such an involution can be shown in three steps.
1) The unitary group $U(n)$ is the intersection of the symplectic group $Sp(2n)$ and of the orthogonal group $O(2n)$.
2) Consider the map $i : Gl(2n) \to Gl(2n)$ given by $i(M) = (M^T)^{-1}$. Since the operations of taking transpose and inverse of a matrices are involutions which commute, we deduce that $i$ is an involution. By definition, the fixed point set of $i$ is $O(2n)$.
3) Since $Sp(2n)$ is invariant under transposition and inversion, $i' := \left. i \right|_{Sp(2n)} : Sp(2n) \to Sp(2n)$ is a well-defined involution whose fixed point set is $Sp(2n) \cap O(2n) = U(n)$.
Now, this involution is not necessarily unique. For instance, for any fixed matrix $U \in U(n) \subset Sp(2n)$, the map $U \circ i' \circ U^{-1}$ is an involution on $Sp(2n)$ which leaves $U(n)$ invariant. However, elements of $U(n)$ are not all fixed by such a map, so $i'$ would still be my bet for what $\theta$ is.
EDIT: Jason DeVito's answer also suggested another solution in the case of $U(n) \subset Sp(2n)$. The complex general linear group $Gl(n, \mathbb{C})$ embeds into the real general linear group $Gl(2n, \mathbb{R})$ as the set of those matrices $M$ which satisfy $MJ_0 = J_0M$ where
$$ J_0 = \left( \begin{array}{cc} 0_{n \times n} & - Id_{n \times n} \\ Id_{n \times n} & 0_{n \times n} \end{array} \right) .$$
In fact, under the identification 
$$\mathbb{C}^n \simeq \mathbb{R}^{2n} : (z_1 = x_1 + iy_1, \dots, z_n = x_n + iy_n) \mapsto (x_1, \dots, x_n, y_1, \dots, y_n),$$
complex multiplication by $i$ in $\mathbb{C}^{2n}$ corresponds to matrix multiplication by $J_0$ in $\mathrm{R}^{2n}$ and $Gl(n, \mathbb{C})$ thus really corresponds to matrices which commute with the complex structure $J_0$. The (noncompact) symplectic group $Sp(2n) \subset Gl(2n, \mathbb{R})$ consists in those matrices $M$ which satisfy $M^TJ_0M = J_0$. Note that $J_0 \in Gl(n, \mathbb{C}) \cap Sp(2n)$. It turns out that the same embedding of $U(n)$ inside $Sp(2n)$ I considered above corresponds to the intersection $Gl(n, \mathbb{C}) \cap Sp(2n)$. Hence another involution $\theta : Sp(2n) \to Sp(2n)$ whose fixed point set is $U(n)$ is given by conjugation by $J_0$ (this is reminiscent of Jason DeVito's $iI$). But in fact, this is the same involution as the one I considered above: indeed, for $M \in Sp(2n)$, we have $$J_0MJ_0^{-1} = (M^T)^{-1}(M^TJ_0M)J_0^{-1} = (M^T)^{-1}(J_0)J_0^{-1} = (M^T)^{-1}.$$
A: Just take $\theta$ to be conjugation by $iI$, where $I$ is the $n\times n$ identity matrix and $i\in \mathbb{C}$.  Then $\theta$ is an involution since $\theta^2$ corresponds to conjugation by $i^2 I = -I$.  Of course, since $-I\in Z(Sp(n))$, conjugation by $-I$ is just another representation of the trivial map.
Now, note that if $A\in U(n)$ (so the entries of $A$ are complex numbers), then $iA = Ai$, owing to the fact that complex numbers are commutative.
Thus, $\theta(A) = iA(-i) = -i^2 A = A$, so $\theta$ fixes $U(n)$ pointwise.
On the other hand, if $B\in Sp(n)\setminus U(n)$, then $B$ must have some entry (say, in the $(s,t)$ slot) with non-trivial $j$ or non-trivial $k$-component.  That is, $B_{st} = b_0 + b_i + b_j + b_k$ with either $b_j\neq 0$ or $b_k\neq 0$.
Then $\theta(B)_{st} = i B_{st} (-i) = b_0 + b_i - b_j - b_k \neq B_{st}$.  So $B$ is NOT fixed by $\theta$.
