Does a zero symmetrized gradient imply a constant function?

Question

Let $\Omega\subset \mathbb{R}^d$ be an open bounded domain. Let $\vec{u}\colon \Omega \to \mathbb{R}^d$ be a vector valued function on $\Omega$. Suppose I know that $\vec{\nabla} \vec{u} + (\vec{\nabla} \vec{u})^\mathrm{T} = 0$ everywhere in $\Omega$, does this mean that $\vec{u}$ is a constant?

Answer 1

Let $d=2$, $(u^1,u^2)=(x_2,-x_1).$