For any function $f$, could one always define an operator $\circ$ so that $f(a\circ b)=f(a)\circ f(b)$? Given any function $f$, I'm interested in knowing whether one could always define an operator $\circ$ (that is, a two place function) such that $$f(a\circ b)=f(a)\circ f(b)$$.
Now, if there exists some element $c$ in the domain of $f$ such that $f(c)=c$ then one could set $a\circ b=c$ for all $a, b$. Yet this is quite trivial. I was wondering if such an operator could always be defined in some non-trivial manner.
 A: Well, I don't know what you consider "trivial", but such a binary operation does always exist: you can always use the binary operation $x\circ y=x$.  Then $f(a\circ b)=f(a)=f(a)\circ f(b)$.  Similarly, $x\circ y=y$ would also work.
Note moreover that given any such binary operation $\circ$, another such binary operation is $x*y=f(x\circ y)$, since $f(a*b)=f(f(a\circ b))=f(f(a)\circ f(b))=f(a)*f(b)$.
In general, there may not be any examples besides these (that is, every example comes from either $x\circ y=x$ or $x\circ y=y$ by composing with $f$ some number of times).  This is true rather vacuously if $f:X\to X$ where $X$ has at most one element.  Less vacuously, if $f:\{0,1\}\to\{0,1\}$ is given by $f(0)=1$, $f(1)=0$, some simple casework shows that the only binary operations that work are $x\circ y=x$, $x\circ y=f(x)$, $x\circ y=y$, and $x\circ y=f(y)$.  (Sketch of proof: composing with $f$ if necessary, we may assume $0\circ 0=0$, which then implies $1\circ 1=1$.  Then by symmetry we may assume $0\circ 1=0$, which implies $1\circ 0=1$ and so we have $x\circ y=x$.)
