Yes. Actually, this was part of my first answer to this question, but this was a digression there (and I also posted there another answer to the same question which addressed it and was accepted). So I'm copying this digression here and will delete the initial answer to that question to avoid a duplicate.
Fact. For every set $X$ there exists $f\in X^X$ whose centralizer in $\mathrm{Sym}(X)$ is reduced to $\{\mathrm{id}_X\}$
It relies on the following second fact: there exists (for $X\neq\emptyset$) a rooted tree structure on $X$ whose automorphism group is trivial. Indeed, granting this, and denoting $v_0$ the root, for a vertex $v$ define $f(v)$ as $v_0$ if $v_0=v$, and as the unique vertex in $[v_0,v]$ at distance 1 to $v$ otherwise. Then $f\in X^X$ and its centralizer in $\mathrm{Sym}(X)$ is the automorphism group of the corresponding rooted tree, which is reduced to $\{\mathrm{id}_X\}$.
To prove the second fact, if $X$ is finite just take a linear tree rooted at an extremal vertex. If $X$ is infinite, by an elementary but very tricky argument (see this answer by user "bof"), there actually exist for every infinite cardinal $\kappa$, $2^{\kappa}$ pairwise non-isomorphic trees of cardinal $\kappa$ each with trivial automorphism group. [Interestingly the induction really requires proving that there are $>\kappa$ such trees, and not only a single one.]