Is it possible to argue that $\nabla F(x^{*})\neq \textbf{0}$? Let $F:\mathbb{R}^{n}\rightarrow \mathbb{R}$ be differentiable. For every affine subset $L$ of $\mathbb{R}^{n}$ of the form
\begin{eqnarray}
L=\{{x}\in \mathbb{R}^{n}:A{x}=b\}
\end{eqnarray}
for some $m\times n$ matrix $A$ having full rank and $b\in \mathbb{R}^{m}$, with $m\leq n$, it is known that $F$ attains a $\textbf{unique minimum}$ (let's call it $x^{*}$) over $L$. Further, it is known that $F$ attains a unique minimum on $\mathbb{R}^{n}$ itself. Let's call this point as $x^{0}$.
Given $L\subset \mathbb{R}^{n}$ of the above form satisfying


*

*$x^{0}\notin L$, and

*$x^{*}\in L$ is the unique point where $F$ attains a minimum in $L$, 


we know, from Lagrange's theorem, that there exists $\lambda^{*}\in \mathbb{R}^{m}$ such that
\begin{eqnarray}
\nabla F(x^{*})=A^{T}\lambda^{*}.
\end{eqnarray} 
Is it possible to argue that $\nabla F(x^{*})=A^{T}\lambda^{*}\neq \textbf{0}$, the all-zero vector? If so, can someone provide a proof of the same? 
Since $F$ attains a global minimum at $x^{0}$, it is clear that 
\begin{eqnarray}
\nabla F(x^{0})=\textbf{0}.
\end{eqnarray}
Since $x^{0}$ is the global minimum, my question is if it is possible that $\nabla F(x^{*})=\textbf{0}$ for $x^{*}\neq x^{0}$ that is the unique point of minimum in $L$.
Nothing more is known about the function $F$, except that it is differentiable. Can imposing more constraints on $F$ (such as requiring $F$ to be strictly convex) pave way for arguing that $\nabla F(x^{*})$ should be a non-zero vector?

Add 1: If we are told that $\nabla F(x^{*})=\textbf{0}$ for some $x^{*}\in \mathbb{R}^{n}$, then we may not be able to conclude if $x^{*}$ is a point of minimum or maximum (or even saddle). However, knowing apriori the fact that $x^{*}$ is the unique point in $L$ where $F$ attains a minimum, and that $x^{0}$ is the unique point in $\mathbb{R}^{n}$ where $F$ attains a minimum, isn't it reasonable to say $\nabla F(x^{*})\neq \textbf{0}$? For, if it were $\textbf{0}$, there would be two points (namely $x^{*}$ and $x^{0}$) where $F$ attains minima in $\mathbb{R}^{n}$? This is just a thought.

It is possible that $\nabla F(\mathbf x^*)=0$ for $\mathbf x^*\ne \mathbf x^0$.
Example: take $f(t)=(t-1)^3$ and consider $F(x,y)=f(x^2+y^2)$. Basically, the graph of $F$ is the rotation of the following curve

The level sets of $F$ are circles, hence, the minimum on any line is unique (intersections of circles and tangent lines are unique). However, the equation
$$
\nabla F(x,y)=2 f'(x^2+y^2)\begin{bmatrix}x\\y\end{bmatrix}=0
$$
has several solutions: the origin (the global minimum of $F$) and the circle $x^2+y^2=1$ where $f'=0$. Hence, taking $L=\{y=1\}$, for example, will give $\mathbf x^*=(0,1)\ne \mathbf x^0=(0,0)$ with $\nabla F(\mathbf x^*)=0$.
The condition that $F$ attains unique minimum on all linear manifolds is equivalent to the level subsets $\{F(\mathbf x)\le C\}$ being strictly convex, that makes $F$ necessarily strictly quasiconvex. If we strengthen it to be pseudoconvex (or, in particular, convex) then 
$$
\nabla F(a)=0\quad\Rightarrow\quad a\text{ is the global minimum}
$$
and by uniqueness it can happen only at $x^0$.
