TITLE: "Anti" fixed point property
QUESTION [19 upvotes]: Let $(X,\tau)$ be a topological space. If $f:X\to X$ is continuous, we say $x\in X$ is a fixed point if $f(x) = x$.
The space $(X,\tau)$ is said to have the anti fixed point property (AFPP) if the only continuous maps $f:X\to X$ with fixed points are the identity map $\text{id}_X:X\to X$, and the constant maps.
The only continuous self-maps of strongly rigid spaces are the identity and the constant maps, so they trivially have the AFPP.

Q. Assume that $(X,\tau)$ is Hausdorff and it has the AFPP. Does this imply that $(X,\tau)$ is strongly rigid?

REPLY [5 votes]: There is a ZFC example based on the so called Cook continuum
(this is a compact metric space with a very rigid structure with respect to selfmaps).
Let $C$ be a Cook continuum and $a, b$ two distinct points of $C$. Let $X$ be the quotient of $C \times \lbrace 0,1 \rbrace$, where the point $(a,0)$ is identified with $(b,1)$ and $(a,1)$ is identified with $(b,0)$. 
Claim: Such a space $X$ has AFPP, but it is not strongly rigid.
There is a nonconstant homeomorphism $h$ which sends the points $(c,0)$ to $(c,1)$ and the points $(c,1)$ to $(c,0)$ for every $c$ from $C$. This $h$ is well defined on the quotient space $X$.
The proof that $X$ has AFPP is using the properties of Cook continuum...
For similar constructions using Cook continuum see the book of Pultr and Trnkova: Combinatorial, algebraic and topological representations of groups, semigroups and categories.