The stable commutator length of a relator is not a one-relator group invariant

Let $S,S'$ be sets and let $r\in F(S)'\setminus\{e\}$, $r'\in F(S')'\setminus\{e\}$ be relators with $\langle {S} \; | \; {r} \rangle \cong\langle{S'} \;| \; {r'}\rangle$. Does this imply that $\text{scl}_S r=\text{scl}_{S'}r'$?

A negative answer to Heuer and Löh's question: the isomorphism type of a one-relator group $\langle S \mid r\rangle$ does not determine $\mathrm{scl}_S(r)$. The witnesses are $r=\mathtt{aabABabABBAbaabABBAb}$ and $r'=\mathtt{aabABabABabABBAbaBAb}$, both length 20 in $F_2'$, with $\langle a,b \mid r\rangle\cong\langle a,b\mid r'\rangle$ but $\mathrm{scl}(r)=1$ against $\mathrm{scl}(r')=1/2$. The mechanism is what made the search finite: $\mathrm{scl}$ is an $\operatorname{Aut}(F_2)$-invariant, so a pair in *different* orbits whose one-relator groups happen to be isomorphic has its two scl values unconstrained by each other. The search was for that configuration among words of length at most 20. Scope: it settles the question as posed and nothing wider. It does not say which invariants do determine scl, and this is a single pair rather than a construction giving arbitrary gaps.