= Solution
For every admissible <transport map> $T$, the <pushforward measure> condition gives $T(x)\in[1,2]$ for $\mu$-almost every $x\in[0,1]$. Hence $|T(x)-x|=T(x)-x$, and
$$
\int_0^1(T(x)-x)\,dx
=\int_1^2 y\,dy-\int_0^1x\,dx=1.
$$
Apply the <Jensen inequality> to the <convex function> $h$ and the <probability measure> $\mu$:
$$
\mathbb M(T)=\int_0^1h(T(x)-x)\,dx\geq h\left(\int_0^1(T(x)-x)\,dx\right)=h(1).
$$
Translation by one sends the source <uniform distribution> to the target <uniform distribution>, so $T^\dagger(x)=x+1$ is admissible. Its displacement is constantly one, giving
$$
\boxed{T^\dagger(x)=x+1,\qquad\min_{T_\#\mu=\nu}\mathbb M(T)=h(1).}
$$
No monotonicity assumption on $h$ is needed, because all displacements have the same sign. The same argument with a <transport plan> also gives the identical minimum for the <Kantorovich optimal transport problem>.
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