Let $X$ be a contractible compact [edit: locally connected] topological space
(Hausdorff and second countable). Let $f\colon X\to X$
be a continuous map. Then (I suppose) $f$ has a fixed
point. Personally, I cannot think of a better generalization
of Brouwer's fixed point theorem, but is it true?
[Math] The generalization of Brouwer’s fixed point theorem
gn.general-topology
Related Solutions
In my experience it is worth considering variants of Lawvere fixed point theorem. In the present case, I would split things up as follows, in order to circumvent the non-constructive nature of Brouwer's fixed point theorem.
Also, let me point out that we need not worry about exponentials too much, even though they do not exist in the category of topological spaces, unless the exponent is nice enough. We can move over to a cartesian-closed subcategory, such as teh compactly generated spaces, or to a cartesian closed supcategory, such as equilogical spaces.
Theorem: [Approximate Lawvere] Suppose $(B, d)$ is a metric space and $e : A \to B^A$ is a continuous map, such that for every continuous map $g : A \to B$ and $\epsilon > 0$ there is $a \in A$ such that $d(e(a)(a), g(a)) < \epsilon$. Then every continuous map $f : B \to B$ has approximate fixed points: for every $\epsilon > 0$ there is $b \in B$ such that $d(b, f(b)) < \epsilon$.
Proof. Given any $f$ and $\epsilon$ consider the map $g(a) = f(e(a)(a))$. there is $a \in A$ such that $d(e(a)(a), g(a)) < \epsilon$ and then $b = e(a)(a)$is an $\epsilon$-approximate fixed point of $f$. QED.
One way to use the theorem is via the sup metric (allowing infinite distance):
Corollary: If $e : A \to B^A$ has a dense image in the sup metric on $B^A$ then every endomap on $B$ has approximate fixed points.
Suppose we could apply the previous theorem to the closed ball $D^n$. Then we would know (constructively!) that every endomap on $D^n$ has approximate fixed points. then we just have another easy step, which contains all the classical reasoning needed:
Theorem: Suppose $X$ is compact and $f : X \to X$ has an $\epsilon$-approximate fixed point for every $\epsilon > 0$. Then $f$ has a fixed point.
Proof. By countable choice, for every $n$ there is $x_n \in X$ such that $d(x_n, f(x_n)) < 1/n$. Because $X$ is compact, $x_n$ has a subsequence converging to some $y \in X$. It is now easy to see that $y$ is a fixed point of $f$. QED.
But I do not see how to apply the approximate Lawvere to the closed ball, if that is even possible.
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.
Best Answer
No. I believe the first counterexample is from:
Kinoshita, S. On Some Contractible Continua without Fixed Point Property. Fund. Math. 40 (1953), 96-98
which I unfortunately can't find online. Kinoshita's example is described on page 127 in this excellent article by Bing, however.
EDIT: The question has been revised to add the local connectivity condition; as stated, I think the question is open. If "contractible" is replaced with "acyclic" there are counterexamples dating back to Borsuk, referenced e.g. here; Borsuk's paper, which I can't find online, is:
K. Borsuk, Sur un continua acyclique qui se laisse transformer topologiquement en lui-meme sans points invariants, Fund. Math. 24 (1935), pp. 51-58.
This suggests that if the OP's conjecture is true, it is unlikely to be open to homological attacks.