Recall that every automatic sequence forms a morphic word. Given an automatic sequence, how can one construct a monoid endomorphism to define the corresponding morphic word?
For example, consider the integer sequence $$ \left( \binom{3n}{n} (\text{mod 2}) : n \in \mathbb{N}_0 \right) = (1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, \cdots) $$ given by sequence A085357 in the On-Line Encyclopedia of Integer Sequences (OEIS). It can be shown that this sequence is $2$-automatic. But it is not obvious as to how to construct a nontrivial monoid endomorphism $$\phi\colon \{ 0, 1 \}^{\ast} \to \{ 0, 1 \}^{\ast}$$ such that the word $1110110011100000111011000...$ is fixed by this morphism.
Is there a general way to find the endomorphism rules to define the morphic word corresponding to an automatic sequence? What is a non-identity endomorphism on $\{ 0, 1 \}^{\ast}$ that fixes the infinite word given by the OEIS sequence A085357?
You are mistakenly assuming the word is purely morphic.
A morphic word is the image of a purely morphic word by some morphism, i.e. there could be secretly different kinds of $0$ and different kinds of $1$ in the purely morphic version of the word, whose difference then get erased by the end morphism.
Since the word is supposedly $2$-automatic, to distinguish between possible variants of what look likes the same letter, you not only need to look at who that letter is, but also at what letters it generates, so its whole subtree : if you are looking at the letter at position $k$, you want to also look at the positions $k,2k,2k+1,4k,\ldots,4k+3,8k,\ldots,8k+7,\ldots$
The number of letters needed in the purely morphic version of the word is the number of different subtrees you can observe.
When you do this you realize that there are two kinds of "$1$"s in your $2$-automatic word: the first kind whose subtree is $1,11,1110,\ldots$ and the second kind whose subtree is $1,10,1100,\ldots$
With the subtree $0,00,0000,\ldots$ this gives you $3$ distinct subtrees (so far), which suggests the following :
Consider the alphabet $\{0,1,\color{red}{1}\}$, and the endomorphism given by
$0 \mapsto 00 ; 1 \mapsto 1\color{red}{1} ; \color{red}{1} \mapsto 10$
Starting with $1$, this gives you the purely morphic word
$1\color{red}{1}101\color{red}{1}001\color{red}{1}1000001\color{red}{1}101\color{red}{1}0000000000\ldots$
After doing the morphism $\color{red}{1} \mapsto 1$, you recover your morphic word
$11101100111000001110110000000000\ldots$
So this endomorphism could be the one you are looking for.