Do $\boldsymbol{\nabla}\cdot \mathbf E=0$ and $\boldsymbol{\nabla}\times\mathbf E=\boldsymbol 0$ imply uniform $\mathbf{E}$? Do the equations
$$\:\boldsymbol{\nabla}\cdot\mathbf{E}=0 \qquad \boldsymbol{\nabla}\times\mathbf{E}=\boldsymbol 0\:$$
imply that the vector field $\mathbf{E}$ is uniform? I think yes, since by Helmholtz Theorem the vector field is uniquely determined and only a uniform vector field satisfies these two conditions?
But then also: $\operatorname{rot} \mathbf{E} = 0$ implies $\mathbf{E} = \operatorname{grad}\Phi$
and $\operatorname{div}(\operatorname{grad}\Phi) = 0$ by Laplace equation. So the solution of the Laplace equation always yields a uniform electric field?
 A: Your second line of thinking is closer to the truth.
It is correct that the gradient of any harmonic function (i.e., a function satisfying $\nabla^2 \phi = 0$) yields a divergence-free and curl-free vector field.  Moreover, the potential for any uniform field $\vec{E}$ is $\phi = - \vec{E} \cdot \vec{r}$, which (it can be shown) is a harmonic function.  So it is true that all uniform vector fields can be derived from a harmonic potential.
However, there are also non-uniform vector fields that derive from harmonic potentials.  For example, $\phi = \frac12 k(x^2 - y^2)$ is a harmonic function, and it leads to a vector field of $\vec{E} = k(-x \hat{\imath} + y \hat{\jmath})$, which is not uniform.
The reason this seems to be at odds with the Helmholtz Theorem is that the  Helmholtz Theorem requires a set of boundary conditions to uniquely determine a vector field, along with its divergence and curl.  In particular, if we require that $\vec{E} \to 0$ as $|\vec{r}| \to \infty$, then the Helmholtz theorem tells us that the only divergence- and curl-free vector field that satisfies this is $\vec{E} = 0$ everywhere.1  As we have seen above, however, there are many other divergence- and curl-free vector fields which are possible if we do not require that $\vec{E} \to 0$ at infinity.

1 Note that Laplace's equation also requires a set of boundary conditions to uniquely determine a solution.  In particular, if we require that $\nabla^2 \phi = 0$ and $\phi \to \text{constant}$ as $|\vec{r}| \to \infty$, then we must have $\phi = \text{constant}$ everywhere.  This is consistent with the fact that the only divergence- and curl-free vector field that vanishes at infinity is the zero vector field.
A: Take $$\vec{E}(x,y,z) = (yz,xz,xy)^t\:.$$
It holds $\nabla \cdot \vec{E}=0$ together with $\nabla \times \vec{E}=\vec{0}$, but the field is not uniform.
The point is that $$\Delta \phi(x,y,z)=0$$  has infinitely many solutions depending on the boundary conditions.
In the case above $\phi(x,y,z) = xyz$ and $\vec{E}= \nabla \phi$ is the gradient of that $\phi$.
A: Not necessarily. Consider the vector field:
\begin{equation}
\vec{E}(x,y,z)
=\begin{pmatrix}
x \\
-y \\
0
\end{pmatrix}
\end{equation}
with $\vec\nabla\cdot\vec{E}=0$ and $\vec\nabla\times\vec{E}=\vec{0}$.
