Commuting functions and fixpoints Let $X$ be an infinite set and let $\text{End}(X)$ be the set of all functions $f:X\to X$. For $f\in\text{End}(X)$ let $$\text{fix}(f) = \{x\in X: x = f(x)\},$$ and $$\text{Com}(f) = \{g\in\text{End}(X): g\circ f = f \circ g\}.$$
It is not hard to prove that if $2^{|\text{fix}(f)|} > |X|$ then $|\text{Com}(f)|> |X|$. Does the converse hold?
EDIT: I forgot to include the condtion that $X$ is an infinite set - apologies.
 A: The answer is no, even if you consider only injective functions. 
Consider any infinite cardinal $\kappa$, and let
$X=\omega\times\kappa$ be the disjoint union of $\kappa$ many
copies of $\omega$. Let $f:X\to X$ be the function that shifts
within each copy separately. So $f$ has no fixed points. But
meanwhile, for any subset $A$ of the copies of $\omega$, let $g_A$ be the
function that shifts within the copies of $A$, but is the identity on the copies of $\omega$ not in $A$. So every $g_A$ commutes with $f$, and there are
$2^\kappa$ many such $g_{A}$. So this is a counterexample to the
converse implication, since $|\text{Com}(f)|=2^{\kappa}>\kappa$ but $2^{|\text{fix}(f)|}=1<\kappa$.
A: Let $n\ge4 $ and suppose that $|X|=n!-1 >2^n$, and that $f$ is a permutation of $X$ with exactly $n$ fixed points. Any permutation $g$ that permutes the fixed points of $f$ and fixes the non-fixed points of $f$ certainly commutes with $f$. Thus $|\mathrm{Com}(f)|\ge n!> |X| >2^{|\mathrm{fix(f)}|}$.
A: Let $X=\mathbf{N}$, $f:X\rightarrow X$ defined by $f(n)=0$ if $n$ even and $f(n)=1$ if $n$ odd. Then $f$ commutes with all elements of the set \begin{multline}
G:=\{g:X\rightarrow X\colon g(0)=0, g(1)=1, g(2k+1)=2m+1, g(2l)=2n, \\k,l,m,n\in X, k,l>0\}.
\end{multline}
Obviously, $|G|=2^{|X|}$, thus $|\text{Com}(f)|\geq |G|=2^{|X|}>|X|>2^{|\text{fix}(f)|}$.
