= Solution
Construct the <conjugacy of positive real-line bumps> using $u=v=\alpha$. Since both $\alpha f$ and $\alpha g$ are greater than $\beta$, choose the fundamental-interval map to be identity on $[\alpha,\beta]$ and then increase from $\beta$ to the respective right endpoints. Equivariant extension along $f$ and $g$ gives a conjugator on their supports fixing $[\alpha,\beta]$. Extend it over the remaining portions of $J=[\gamma,\gamma h]$, fixing the endpoints of $J$; both support intervals lie strictly inside $J$. Call the resulting interval bijection $b_0$. It satisfies $fb_0=b_0g$ on $J$.
The point $\gamma$ lies in a positive orbital of $h$, because $\gamma h>\gamma$. Its half-open cells $J h^n$, $n\in\mathbb Z$, partition that orbital. Extend by
$$
(th^n)b=(tb_0)h^n,\qquad \gamma\le t<\gamma h,
$$
and set $b$ to identity outside the orbital. Since $b_0$ fixes the cell endpoints, this is an <order automorphism> and satisfies $bh=hb$. Each cell is preserved; $f$ and $g$ are identity on all cells other than the original one, so $fb=bg$ holds everywhere. Finally $b$ fixes $[\alpha,\beta]$ pointwise, and $a$ is supported inside that interval. Hence $ba=ab$. Thus
$$
\boxed{b^{-1}fb=g,\qquad ba=ab,\qquad bh=hb}.
$$
This is <conjugating positive bumps while preserving a fundamental interval>; merely choosing an arbitrary conjugator would not ensure either commutation relation.
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