Suppose that $F:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is $L$-Lipschitz continuous (for some $L>0$) and monotone, i.e., \begin{align} (\forall x,y \in \mathbb{R}^{n}) \quad \| F(x) - F(y) \| \leq L \| x-y \|, \end{align} and \begin{align} (\forall x,y \in \mathbb{R}^{n}) \quad \langle F(x) - F(y), x-y \rangle \geq 0, \end{align} respectively.
Now, suppose that $\gamma \in (0,1/L)$ and $x\in \mathbb{R}^{n}$ is such that \begin{align} F(x) = F(x-\gamma F(x)) \end{align} and $x^{\star}\in \mathbb{R}^{n}$ such that $F(x^{\star})=0$.
Question: Does this imply that $F(x) = 0$? If not, what is a counterexample?
What I have: I'm unsure if this is true, but I don't have a counterexample yet.
On the other hand, using monotonicity, I get that \begin{align} \gamma \|F(x)\|^2 \leq \langle F(x),x-x^{\star} \rangle, \end{align} and by Lipschitz continuity, I get that \begin{align} \|F(x)\| \leq L \|x-\gamma F(x) - x^{\star}\|. \end{align} However, I cannot show that $F(x)=0$ with these inequalities.
Edit: Consider $F:\mathbb{R}\to\mathbb{R}$ such that \begin{align} F(x) = \begin{cases} x+1 & \text{if } x<-1 \\ 0 & \text{if } -1\leq x< 1 \\ x-1 & \text{if } 1 \leq x < 2 \\ 1 & \text{if } 2 \leq x. \end{cases} \end{align} Then $F$ is $1$-Lipschitz continuous and monotone. Clearly $F(0) = 0$. Let $x=4$ and $\gamma = 1/2$. Then \begin{align} F(x) &= F(4) = 1 \end{align} and \begin{align} F(x-\gamma F(x)) &= F(3.5) = 1 \end{align}