We can show that $U(n) \subset SO(2n)$. For example, we can see for a $n$-dimensional complex $\mathbb{C}$ vector: $$ Z_j=X_j +i Y_j $$ for $j=1,\dots, n$ being acted by the rank-$n$ $U(n)$ matrix representation, we can map $$Z_j=X_j +i Y_j \text{ acted by } U(n) \to (X_1, \dots, X_j, \dots, X_n, Y_1, \dots, Y_j, \dots, Y_n) \text{ acted by } SO(n).$$ The $2n$-dimensional real $\mathbb{R}$ vector $(X_1, \dots, X_j, \dots, X_n, Y_1, \dots, Y_j, \dots, Y_n)$ can be acted (or transformed) by the rank-$2n$ $SO(2n)$ matrix representation.
We see that the invariance of $U(n)$ from two vectors $Z'$ and $Z$ $$ Z'^* \cdot Z\equiv (\dots,X_j' -i Y_j',\dots)\cdot(\dots,X_j +i Y_j,\dots) =\sum_{j} (X_j'X_j + Y_j' Y_j)+ i (- Y_j'X_j + X_j' Y_j)= \text{invariant}, $$ contains both the constraints from the $SO(2n)$ invariant: $$ \sum_{j} (X_j'X_j + Y_j' Y_j)= \text{invariant}, $$ but also $$ \sum_{j} (- Y_j'X_j + X_j' Y_j)= \text{invariant}. $$
My puzzle: Can we prove that $$U(n) \subset Spin(2n)?$$ or prove $$U(n) \not \subset Spin(2n)?$$ Note $U(n)=(SU(n)\times U(1))/(\mathbb{Z}/n\mathbb{Z})$.