The same triple can accept for one machine and reject for another

Both machines remove the same set $\{A,R\}$ from $Q$. Their nonhalting domain is therefore exactly $\{(s,a),(s,\bigsqcup )\}$, and the two displayed entries provide a unique correctly typed output at each pair. In $M$ the designated triple is $(s,A,R)$, and in $N$ it is $(s,R,A)$; each has distinct entries. Their finite sets and blank exclusions are the same, so both tuples satisfy all the requirements.

The blank tape has support $\{i\in \mathbb{N}:\bigsqcup \mathrm{\ne }\bigsqcup \}=\mathrm{\varnothing }$. It is consequently a tape for both machines, and $A\in Q$ and $0\in \mathbb{N}$ make $C$ a configuration of both. The shared state, head, and tape components require no change when the designations are interchanged.

For $M$ the accepting equality is $A=q_{\mathrm{a}\mathrm{c}\mathrm{c}}^M=A$, which is true, whereas the rejecting equality is $A=q_{\mathrm{r}\mathrm{e}\mathrm{j}}^M=R$, which is false. Thus $C$ is accepting and not rejecting for $M$.

For $N$ the rejecting equality is $A=q_{\mathrm{r}\mathrm{e}\mathrm{j}}^N=A$, which is true, whereas the accepting equality is $A=q_{\mathrm{a}\mathrm{c}\mathrm{c}}^N=R$, which is false. Thus $C$ is rejecting and not accepting for $N$. It is halting for both machines, but the named outcome depends on the machine's designations, as asserted.