Let $X = \{0,1\}^{\mathbb{Z}}$, and $\sigma$ the left shift. Let $\mathcal{Q}_n(X)$ be the set of $\sigma$-orbits of cardinality $n$, i.e. the orbits of points of least period $n$ in $X$. If $f \in \mathrm{Aut}(X)$, then $f$ naturally acts on $\mathcal{Q}_n$ i.e. orbits are mapped to orbits. The *signs* of $f$ are defined as the signs of these permutations, i.e. $s_n(f) = -1$ if $f$ is an odd permutation on $\mathcal{Q}_n$, and $s_n(f) = 1$ otherwise.

The *gyration numbers* of $f$ are defined as follows: For each $\gamma \in \mathcal{Q}_n(X)$, pick a representative $x_\gamma \in \gamma$. Let $k_\gamma$ be such that $f(x_\gamma) = \sigma^{k_\gamma}(x_{f(\gamma)})$. Then $g_n(f) = \sum_{\gamma \in \mathcal{Q}_n(X)} k_\gamma$. It is easy to check that this does not depend on the choice of representatives, since if you shift the representative you decrease one $k$-value and increase another.

Finally we say $f$ satisfies the *sign-gyration compatibility condition* if for all $q$ odd and all $m$, we have
$g_{2^m q}(f) = \left\{\begin{array}{ll}
0 & \mathrm{if} \prod_{j = 0}^{m-1} s_{2^j q}(\phi) = 1 \\
2^{m-1} q & \mathrm{if} \prod_{j = 0}^{m-1} s_{2^j q}(\phi) = -1
\end{array}\right.$

All we need to notice in this formula is that $g_n(f)$ is always equal to $0$ or $n/2$ (when this condition holds).

Now, according to [1], [2] proves that there is a shift-commuting permutation of the points of least period 6 of $X$, which is not the restriction of an automorphism of $X$. This solves your question in the negative, taking $A = B$ the set of points of least period $6$.

I cannot access [2], but it seems that this follows immediately from a (to me) better-known result also from [2] (though I don't know the proof of this one either). Namely, again according to [1], Kim and Roush prove in [2] that the SGCC holds for every inert automorphism of every mixing SFT. Inert automorphisms are the ones in the kernel of the dimension representation, i.e. they act trivially on the dimension group.

I won't define the dimension representation here (see e.g. [1] for the definition), but it is known that the dimension representation of the binary full shift $X$ is isomorphic to $\mathbb{Z}$, and the shift map $\sigma$ generates it.

From the sign-gyration result of [2] it then follows that the gyration numbers for period $6$ are generated by $g_6(\sigma) = 3$ and whatever SGCC allows (i.e. $0$ and $3$), since gyration is a group homomorphism. So we deduce that no automorphism $f$ can satisfy $g_6(f) = 1$. But there is a permutation of the $6$-periodic points which would give $g_6(f) = 1$ for any automorphism extending this, for example the automorphism that rotates the orbit of $000001$, but does not rotate any other orbit.

**References**

[1] *Lind, Douglas; Marcus, Brian*, **An introduction to symbolic dynamics and coding**, Cambridge: Cambridge University Press (ISBN 0-521-55124-2/hbk; 0-521-55900-6/pbk). 484 p. (1995). ZBL1106.37301.

[2] *Kim, K. H.; Roush, F. W.*, On the structure of inert automorphisms of subshifts, PU.M.A., Pure Math. Appl., Ser. B 2, No. 1, 3-22 (1991). ZBL0766.54041.

canalways extend a bijection to an automorphism, just permute symbols. $\endgroup$