Solution $\chi$ of $\chi_k(x)=\chi_x(f_k(x))$ given $f_k$ bijective

Question

Given a family $\{ f_a \}_{a\in H}$ of bijective functions over a set $H$ I need to find a family of functions $\{ \chi_a \}_{a\in H}$ from $H$ to $H$ such that for every fixed $k$ in the family

$$\chi_k (k)=\chi_k(f_k(k))$$

And $\forall k,x \in H$

$$i) \,\, \chi_k(x)=\chi_x(f_k(x))$$

... or maybe we can use $\{ f_a \}_{a\in H}$ to define a binary operation $*_f$ on $H$ in this way:

$$a*_f x:= f_a (x)$$

and we must to find a binary operation $*_\chi$ on $H$ witht this property

$$ib) \,\, x *_\chi k= (k *_f x) *_\chi x$$

1-Only with this assumption and the associativity, is the binary operation $*_\chi$ unique?

2-Can the choice of $*_f$ (the family of bijective fucntion) "force" $*_\chi$ to be the only solution of the equation?

3-If not why? There is some general theorem about the uniqueness of the solution of these kind of equations?

Answer 1

It seems the following.

1. In general, the answer is negative. For instance, if $a*_f x=x$ then Condition (ib) transforms to $x *_\chi k= x *_\chi x$. Then for each function $g:H\to H$ a binary operation $*_\chi$ on $H$ such that $x *_\chi k=g(x)$ for each elements $x,k\in H$ satisfies this condition.
2. If $|H|>1$ then the answer is negative, because each constant operation $*_\chi$ on $H$ satisfies Condition (ib).