A variation of the Ryll-Nardzewski fixed point theorem Is there a fixed-point theorem that implies the following result?

Let $F$ be a nonempty convex set of functions on a discrete group with values in $[0,1]$. Suppose $F$ is invariant with respect to left shifts and closed with respect to the pointwise convergence. Then $F$ contains a constant function.

This statement looks like Ryll-Nardzewski fixed point theorem, but it does not seem to follow from the theorem.
 A: The claim does not hold, as the nice example by Naratuka Ozawa shows.
The purpose of this answer (or rather, extended comment) is to share a related fixed point theorem.
Theorem:
Let $X$ be a non-empty set endowed with two structures:

*

*The structure of a compact convex set, that is a convex structure and a compatible compact topology $T$ such that continuous affine functionals separate the points.

*The structure of a separable metric space $(X,d)$.

Assume the following compatibility of the two structures: $d$-balls in $X$ are measurable with respect to the Borel $\sigma$-algebra associated with $T$.
If $G$ is a group that acts on $X$ preserving both structures - the convex structure, the topology $T$ and the metric $d$ - then $G$ has a fixed point in $X$.

Let me illustrate how the theorem above implies the fixed point theorem of Ryll-Nardzewski: for a non-empty weakly compact subset $C$ in a Banach space, there is a point in $C$ which is fixed by all affine isometries of $C$.
It is enough to show that $C^G$ is non-empty for every countable group of affine isometries $G$. Given such a group $G$, fix a point $c\in C$ and set $X$ to be the norm closure of the convex hull of the orbit $Gc$. Take $T$ to be the weak topology and $d$ to be the norm metric on $X$. We are now in a position to apply the above theorem.

Sketch op the proof:
As above, we assume as we may that $G$ is countable.
We fix a fully supported probability measure $\mu$ on $G$ and denote by $B$ the corresponding Furstenberg-Poisson boundary. Then the $G$ action on $B$ is amenable and metrically ergodic. By amenability and by 1 there exists a measurable, defined a.e, $G$-equivariant map $B\to X$. By the compatibility of 1 and 2, this map is $d$-measurable, thus by metric ergodicity and by 2 this map must be essentially constant. Its essential image is a fixed point.
A: The claim does not hold. Let $F$ be the free group on $\{a,b\}$ and $X⊂F$ be the words whose last letter is $a$ or $b$. Let $\xi=\delta_a+\delta_b−\delta_{a^{-1}}−\delta_{b^{-1}}\in\ell_1(F)$. Then for any $s\in F$, one has $\langle 1_X,s\xi\rangle\geq1$. (To see this, view $F$ as the $4$-regular tree and $s\xi$ as a signed characteristic function of the $1$-neighborhood of $s\in F$.) Hence the pointwise-topology closed convex hull of $F1_X$ in $\ell_\infty(F)$ cannot contain constant functions.
