An easy question that I have never been able to answer.
Suppose we have the CA on $\{ 0,1,2 \}^{\mathbb{N}}$ with local rule given by $f(x,y)=A\_{x,y}$ and $A$ the $3\times 3$ matrix $A=(0,1,2,0,1,2,1,2,0).$ For example $(0,1,2,0,0,0,1,2,0,\ldots)\mapsto (1,2,1,0,0,1,2,1,\ldots),$ It is like a shift if the coordinate before is $0$ or $1$ and $x+1$ mod 3 if not.
My question is, are there any invariant probability measures full supported other that the Haar measure? I have never seen an idea to solve this problem, so any reference is welcome.



Invariant measures for Cellular automata
This problem is easy, but there is a related problem that is not, this is to find a fully supported invariant measure for our CA that is at the same time invariant for the action of the shift different from the Haar measure. Is this possible? 


To solve the problem here we can do the following. Call our CA by $F.$ $F$ is clearly surjective, then by a result by Hedlund every cylinder set has 3 preimages.


We will find an invariant measure for $F$ by defining recursively in $k$ the measure on the cylinder sets with coordinates fixed $0,1,\ldots,k,$ then we will 
extend the measure using Komogorov as usual. Define for $i\in \{0,1,2\}$ define $[i]\_0=\{ i\}\times \{0,1,2\}^{\mathbb{N}}.$


For $k=1,2,\ldots$ suppose $F^{-k}[i]\_0=\{a\_i^{1}(k),a\_i^{2}(k),\ldots,a\_i^{3^{k}}(k)\},$ for example ordered by the lexicographic order. We will define inductively in $k$ the values $\mu(a\_i^{j}(k))\doteq p\_{i}^{j}(k)\in (0,1).$ 


Choose $p\_0^{1}(0),p\_1^{1}(0),p\_2^{1}(0)>0$ such that $p\_0^{1}(0)+p\_1^{1}(0)+p\_2^{1}(0)=1.$ Define $\mu([i]\_0)=p\_i^{1}(0).$ 


Suppose we have defined $\mu$ in $F^{-k}[i]\_0.$ Given $j=1,2,\ldots,3^{k}$ we find $j\_1,j\_2,j\_3$ such that $F(a\_i^{j\_1}(k+1))=F(a\_i^{j\_2}(k+1))=F(a\_i^{j\_3}(k+1))=a\_i^{j}(k)$ (the evaluation of $F$ on cylinder sets has the obvious meaning) choose $p\_i^{j\_1}(k+1),p\_i^{j\_2}(k+1),p\_i^{j\_3}(k+1)>0$ such that $p\_i^{j\_1}(k+1)+p\_i^{j\_2}(k+1)+p\_i^{j\_3}(k+1)=p\_i^{j}(k).$


Clearly we have defined in such a way an $F$-invariant probability measure, however not necessarily shift invariant.


