Let $X$ a compact set and $A\subseteq \mathbb{R}$. Consider a continuous function $f\colon X\times A\to X$ and construct a fixed-point iteration as follows $$ x_{k+1}=f(x_k,a),\quad x_0\in X, a \in A.\quad (\star) $$
My question: If $(\star)$ admits a unique fixed point, denoted by $\mathrm{Fix}(f_a)$, for all $a\in A$, can we conclude that $\mathrm{Fix}(f_a)$ is a continuous function of $a$? What can be said in case $(\star)$ admits a set of fixed points for all $a\in A$?
Thanks for your help.
Comments. This question is different from this one. Indeed, here $f$ is not assumed to be contractive.
Suppose $X$ is metric (with metric $d_X$), and consider the following parameterized family of optimization problems: $$ \max_{x \in X} \, -d_X\big(x, f(x,a)\big) \tag{$\ast$} $$ By construction, $d_X\big(x, f(x,a)\big)$ is continuous in $(x,a)$, and given $X$ is compact, it is straightforward to verify the other conditions of Berge's Theorem of the Maximum.
Consider the argmax correspondence, i.e. the possibly set-valued mapping $\phi: A \rightrightarrows X$ defined via $a \mapsto \{x \in X : f(x,a) =x\}$. By Berge, we obtain that $\phi$ is upper hemicontinuous as a correspondence.
Any upper hemicontinuous correspondence that additionally is singleton-valued is a continuous function. Thus, if each $f(\cdot, a)$ admits a unique fixed point, $\phi$ is a continuous function. If some $f(\cdot, a)$ admit more than one fixed point, $\phi$ will not be singleton-valued, but will instead be upper hemicontinuous as a correspondence (here, equivalent to having a closed graph in $A \times X$).
Note nothing in this requires these fixed points to be attracting under the specified dynamics.