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](https://mathoverflow.net/questions/188729/strongly-rigid-hausdorff-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?
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"Anti" fixed point property
Probably, the discrete $\{0,1\}$ is not the counterexample Dominic van der Zypen expected to see :) 


A more elaborate CH-example of a AFPP but not strongly rigid space was constructed by van Mill:


**Theorem 4.1 ([van Mill, 1983](http://www.ams.org/journals/tran/1983-280-02/S0002-9947-1983-0716833-2/S0002-9947-1983-0716833-2.pdf))**. Under Continuum Hypothesis there exists a non-trivial metrizable separable connected locally connected Boolean topological group $X$ such that each continuous self-map $f:G\to G$ is either a translation or a constant map. 


The space $G$ has AFFP but is not strongly rigid.


**Remark.** In the van Mill's proof the Continuum Hypothesis is used in combination with the following classical result of Sierpiński:


**Theorem (Sierpiński, 1921).** For any countable partition of the unit interval into closed subsets exactly one set of the partition is non-empty.


Motivated by this Sierpiński Theorem we can ask about the smallest infinite cardinality $\acute{\mathfrak n}$ of a partition of the unit interval into closed non-empty subsets. It is clear that $\acute{\mathfrak n}\le\mathfrak c$. The Sierpinski Theorem guarantees that $\omega\_1\le\acute{\mathfrak n}$. This inequality [can be improved](https://mathoverflow.net/questions/285780/is-the-sierpi%C5%84ski-cardinal-acute-mathfrak-n-equal-to-some-known-small-uncou) to $\mathfrak d\le\acute{\mathfrak n}\le\mathfrak c$.


It seems that the proof of van Mill's Theorem actually yields more:


**Theorem**. Under $\acute{\mathfrak n}=\mathfrak c$ (which follows from $\mathfrak d=\mathfrak c$) there exists a non-trivial metrizable separable connected locally connected Boolean topological group $X$ such that each continuous self-map $f:G\to G$ is either a translation or a constant map. 


