= Solution
One useful version is the effective <global Torelli theorem for K3 surfaces>. Let $X,Y$ be complex <K3 surfaces> and let
$$
\phi:H^2(Y,\mathbb Z)\xrightarrow{\sim}H^2(X,\mathbb Z)
$$
preserve the integral cup-product pairing and the Hodge decomposition. Suppose that $\phi$ carries some <Kähler class> of $Y$ to a <Kähler class> of $X$. Then
$$
\boxed{\text{there is a unique isomorphism }f:X\xrightarrow{\sim}Y
\text{ such that }f^*=\phi.}
$$
Conversely every <isomorphism> has these properties. In particular, for projective <K3 surfaces> an integral Hodge isometry carrying an ample polarization to an ample polarization is induced by a unique <isomorphism> of the polarized surfaces. This is the https://www.math.uni-bonn.de/people/huybrech/K3Global.pdf[effective global Torelli statement].
The Kähler-cone condition is necessary if the prescribed isometry itself is to be induced by an <isomorphism>. If $\delta$ is the class of an effective smooth rational $(-2)$-curve, the reflection
$$
r_\delta(x)=x+(x,\delta)\delta
$$
is integral, preserves the pairing, and preserves the Hodge decomposition because $\delta$ has type $(1,1)$. It sends $\delta$ to $-\delta$. An <automorphism> preserves effectivity, whereas $-\delta$ is not effective: it has negative intersection with an ample class. Thus this Hodge isometry is not an <automorphism> pullback. Requiring an isometry to respect the Kähler chamber removes precisely this kind of obstruction.
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