Let $A\in\mathbb{R}^{n\times n}$ be a nonsingular matrix, $b\in\mathbb{R}^n$, and $f\colon\mathbb{R}\to\mathbb{R}$ be an arbitrary function. Consider the following differential equation $$\tag{$\ast$}\label{ast} \dot{x}(t)=-x(t)+f(Ax(t)+b),\ \ x(0)\in\mathbb{R}^n, $$ where, for a vector $x$, $f(x)$ acts component-wise on the entries of $x$, i.e. $[f(x)]_i=f(x_i)$.
Using the change of variable $y(t):=Ax(t)+b$, it seems to me that above differential equation can be equivalently rewritten as $$\tag{$\ast\ast$}\label{astast} \dot{y}(t)=-y(t)+Af(y(t))+b,\ \ y(0)\in\mathbb{R}^n. $$
My questions:
Is it true that \eqref{ast} admits a unique equilibrium (i.e., a unique $\bar{x}\in\mathbb{R}^n$ such that $\bar{x}=f(A\bar{x}+b)$) if and only if \eqref{astast} admits a unique equilibrium?
(In case the answer to question 1 is in the affirmative) Is it true that \eqref{ast} admits a unique locally stable equilibrium if and only if \eqref{astast} admits a unique locally stable equilibrium?