Morphisms generating antipalindromic words

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Problem: Reconstructed statement: over the binary alphabet $\{0,1\}$, let $E(w_1\cdots w_n)=(1-w_n)\cdots(1-w_1)$. A finite word $w$ is an antipalindrome if $E(w)=w$, and an infinite word is antipalindromic if it has infinitely many antipalindromic factors. Conjecture 42 asserts:

If $\varphi:\{0,1\}^*\to\{0,1\}^*$ is a primitive morphism with an antipalindromic fixed point $\mathbf u$, then $\varphi$ or $\varphi^2$ is conjugate to a morphism in $\mathcal A_1\cup\mathcal A_2$.

Here $\mathcal A_1$ consists of morphisms

with $p\neq\varepsilon$ and $s$ antipalindromic, hence all morphisms in $\mathcal A_1$ are uniform.
$\mathcal A_2$ consists of morphisms

where $w\neq\varepsilon$, $k,h\ge 0$, and $\Theta(0)=01,\Theta(1)=10$.

Result: The literal conjecture is false. Define

Then $\varphi$ is primitive, since both images contain both letters. Its fixed point starting in $0$ is

because

This word is antipalindromic: for every $n\ge1$, $(01)^n$ is an antipalindrome.

But $\varphi$ is not conjugate to a morphism in $\mathcal A_1$, since $|\varphi(0)|=2$ and $|\varphi(1)|=4$, while every $\mathcal A_1$-morphism is uniform; conjugacy preserves image lengths.

Nor is $\varphi$ conjugate to a morphism in $\mathcal A_2$. For any $\psi\in\mathcal A_2$,

so the two lengths have ratio a ratio of odd integers. The ratio $2:4=1:2$ is not such a ratio.

The same obstruction applies to $\varphi^2$:

again ratio $1:2$. Thus $\varphi^2$ is also not conjugate to any morphism in $\mathcal A_1\cup\mathcal A_2$.

So Conjecture 42 fails because of a periodic fixed point. The minimal natural repair is to add the hypothesis that the fixed point is aperiodic; that is consistent with the paper’s main theorems, which treat the aperiodic case separately.

Citation: Counterexample above is elementary. Definitions are from Ambrož–Masáková–Pelantová, “Morphisms generating antipalindromic words,” arXiv:1906.06174.