$\mathbb{C}$ is a fixed finite set, a fair chaotic sequence $(c_n \in \mathbb{C})$ is defined such that $\forall c \in \mathbb{C}, \exists n_0 \in \mathbb{N}, n > n_0 \wedge c_n = c$. That means all elements of $\mathbb{C}$ are encountered infinitely often in the sequence $(c_n)$.
I define a function $f: \mathbb{C} \times \mathbb{X} \rightarrow \mathbb{X}$. Given $x_0 \in \mathbb{X}$, $\forall n \geq 0,x_{n+1} = f(c_{n}, x_{n})$ generates a sequence $(x_n)$. $f$ is defined such that for all fair sequence $(c_n)$ as inputs, there exists a common $x \in \mathbb{X}$ such that $\exists n_0\in \mathbb{N}, \forall n \geq n_0, x_n = x$. That means the sequence $(x_n)$ will stabilize on a value no matter what a fair sequence $(c_n)$ is.
My question is how to name this stable value, can we call it a fixpoint? But a fixpoint of a function implies the function takes one input and returns one output, whereas in our case, $f$ takes an extra sequence $(c_n)$ as inputs.
Otherwise, can we call it a stable point of $f$? Could anyone help?
The issue of inputs can be fixed by a slight change of notation. Try writing $f_c(x)$ for $f(c,x)$. This means that instead of a map from $\mathbb C\times X$ to $X$ we think of a finite collection $\{f_c:c\in\mathbb C\}$ of maps of $X$ onto itself. You define $$x_{n+1}=f_{c_n}(x_n)=(f_{c_n}\circ f_{c_{n-1}}\circ \dots f_{c_0})(x_0) $$ If $x$ has the property you described, then $f_c(x)=x$ for all $c\in \mathbb C$. This is expressed by saying that $x$ is a common fixed point of the family $\{f_c:c\in \mathbb C\}$.
Or you can borrow terminology from Markov chains and call $x$ an absorbing state.
By the way, the property is interesting. I'd like to see an example where such $x$ exists for non-obvious reasons.