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.



Commuting functions and fixpoints
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 $|\text{Com}(f)|\geq |G|=2^{|X|}>|X|>2^{|\text{fix}(f)|}$.


