"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). 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 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.
The set $X=\{0, \, 1\}$ endowed with the discrete topology shows that your question has a negative answer.
In fact, the only non-constant self-map of $X$ different from the identity is the non-trivial involution, which is a fixed-point free self-homeomorphism.
Hence $X$ has the AFPP but it is not strongly rigid.
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.