A **synchronized system** is a transitive shift space $X$ which has a **synchronizing block** $v$, that is $v$ is an admissible block for $X$ and whenever $vw$ and $uv$ are admissible blocks in $X$, then $uvw$ is also admissible. 
Let $\Psi\colon X\to Y$ be a conjugacy of synchronized systems. 


Since $\Psi$ is continuous and shift-commuting there is an $n\in\mathbb{N}$, a map from admissible $n$-bloks in $X$ to $1$-blocks in $Y$, denoted $\psi\colon \mathscr{B}\_n(X ) \to
\mathscr{B}\_1(Y)$, and an $m \in\mathbb{Z}$ such that, in standard notation, $(\Psi \circ \sigma^m(x))\_i = \psi(x\_{[i,i+n)})$,
for all $i \in \mathbb{Z}$ and all $x \in X$, cf. Lind and Marcus book. The map $\psi$ extends to maps $\psi^{(k)} \colon \mathscr{B}\_k(X ) \to
\mathscr{B}\_{k-n+1}(Y)$, $k \ge n$, in the natural way. 


Is it true that a $k$-block $v$ ($k\ge n$) is a synchronizing block for $X$ **if and only if** $\psi^{(k)}(v)$ is a synchronizing block for $Y$? 


In general, a factor of a synchronized system may not be synchronized.



Does conjugacy preserve the set of synchronizing blocks?
No! If so, you would get a contradiction by considering the length of the shortest synchronizing blocks of $X$.


Namely, suppose that the synchronizing blocks of $X$ all have length at least $2$. Let $v$ be a synchronizing block of $X$ having length $n\geq 2$. Choose a subshift $Y$ that is conjugate to $X$ via an $n$-block map $\psi$. (For example, $Y$ could be the $n$-th higher block presentation of $X$.) The inverse map from $Y$ to $X$ is induced by another block map $\varphi$, which has window size $m\geq 1$.


If the claim is true, $Y$ must have a synchronizing block $c=\psi(v)$ of length $1$. If $c$ is synchronizing for $Y$, so is every extension of $c$ to an admissible block. In particular, $Y$ has a synchronizing block $u$ of length $m$. Applying the claimed statement again we obtain that $X$ must have a synchronizing block $\varphi(u)$, which has length $1$, hence a contradiction.


