= Solution
Use the closed <genus> $g\geq2$ version of <Teichmüller's uniqueness theorem>. Let $f_0:X\to Y$ be a <Teichmüller map>: for a nonzero <holomorphic quadratic differential> $q$, normalized by $\int_X|q|=1$, and $0<k_0<1$,
$$
\mu_{f_0}=k_0\frac{|q|}{q},\qquad K_0=\frac{1+k_0}{1-k_0}.
$$
The value at a zero of $q$ is irrelevant to the <Beltrami coefficient>, which is defined almost everywhere. For every <quasiconformal map> $f:X\to Y$ preserving <orientation> and <homotopic> to $f_0$, the conclusion is
$$
\boxed{K(f)\geq K_0,\qquad K(f)=K_0\ \Longrightarrow\ f=f_0.}
$$
The normalized $q$ is also uniquely determined when $k_0>0$.
We use the following analytic input, stated with its hypotheses. The <Reich–Strebel inequality> says that if $f_0$ has the displayed <Beltrami coefficient> and a competitor $f$ has the same target and <homotopy class>, then, for $\mu=\mu_f$,
$$
K_0\leq\int_X\frac{|1+\mu q/|q||^2}{1-|\mu|^2}\,|q|.
$$
This is the standard fundamental inequality for an integrable <holomorphic quadratic differential>; it also holds on a finite-type punctured surface with the homotopy fixing the punctures. Its formulation is given in https://userhome.brooklyn.cuny.edu/gardiner/A%20short%20course%20on%20Teichmuller%27s%20theorem.pdf[Gardiner and Hu, §5]. We quote this analytic inequality as the lecture result used in the proof.
Put $k=\|\mu\|_\infty$. Pointwise, away from the isolated zeros of $q$,
$$
\frac{|1+\mu q/|q||^2}{1-|\mu|^2}
\leq\frac{(1+|\mu|)^2}{1-|\mu|^2}
=\frac{1+|\mu|}{1-|\mu|}
\leq\frac{1+k}{1-k}.
$$
Integrating against the <probability density function> $|q|$ proves $K_0\leq K(f)$. If $K(f)=K_0$, all these inequalities are equalities almost everywhere. The strictly increasing last function forces $|\mu|=k_0$ almost everywhere; equality in the <triangle inequality> then forces $\mu q/|q|=k_0$. Therefore
$$
\mu_f=\mu_{f_0}\quad\text{almost everywhere}.
$$
Two <quasiconformal maps> with the same <Beltrami coefficient> differ by postcomposition with a <biholomorphism>, by the local chain rule for the <Beltrami equation>. Consequently $h=f\circ f_0^{-1}$ is a <biholomorphism> from $Y$ to itself <homotopic> to the identity.
For completeness, such a <biholomorphism> is the identity when $g\geq2$. Apply the <uniformization theorem> and choose the lift of the homotopy to $\mathbb H$ starting at the identity. Its endpoint lift $\widetilde h$ commutes with every <deck transformation>. It is a real <Möbius transformation>. The <compact> quotient has no <parabolic Möbius transformations> in its deck group, and freeness excludes <elliptic Möbius transformations>. Two distinct hyperbolic axes exist: a discrete free group preserving just one axis would be cyclic, contradicting the <fundamental group> of a <closed surface> of <genus> at least two. Commutation with two <deck transformations> represented by <hyperbolic Möbius transformations> having distinct axes makes it fix their boundary endpoints individually; there are at least three such endpoints. A <Möbius transformation> fixing three points is the identity. Hence $h=\operatorname{id}$ and $f=f_0$.
If $q_1$ is another unit-area differential for the same map, then $|q_1|/q_1=|q|/q$, so $q_1/q$ is positive real wherever defined. This <meromorphic function> is constant by the <open mapping theorem>, and normalization makes the constant one. Without normalization, positive multiples of $q$ describe the same map. At $k_0=0$, uniqueness of the map still holds in <genus> at least two, but there is no distinguished $q$. In <genus> one, translations supply nontrivial <biholomorphisms> <homotopic> to the identity, so equality determines the map only up to those <biholomorphisms>; the <genus> hypothesis cannot be omitted.
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