= Solution
An <isometry> preserves the <Levi-Civita connection>. To justify this, transport the connection on $X$ by $d\psi$. It is torsion-free and preserves the metric on $Y$, since $\psi$ preserves the metric and Lie brackets. These two properties characterize the <Levi-Civita connection> uniquely: if two such connections differ by $A$, the tensor $a(U,V,W)=\langle A(U,V),W\rangle$ is symmetric in its first two arguments and antisymmetric in its last two. Cycling these symmetries yields $a=-a$, hence $A=0$. Therefore
$$
d\psi(\nabla^X_UV)=\nabla^Y_{d\psi U}(d\psi V).
$$
Applying this along $\gamma$ shows that the covariant acceleration of $\psi\circ\gamma$ is zero, so \b[an <isometry> carries affinely parametrized <geodesics> to <geodesics>].
The same identity and the definition $R(U,V)W=\nabla_U\nabla_VW-\nabla_V\nabla_UW-\nabla_{[U,V]}W$ show that $d\psi$ intertwines the <Riemann curvature tensor>. If $e_1,e_2$ is an <orthonormal> tangent basis, its image is <orthonormal> and
$$
K_Y(\psi(p))=\langle R_Y(d\psi e_1,d\psi e_2)d\psi e_2,d\psi e_1\rangle
=\langle R_X(e_1,e_2)e_2,e_1\rangle=K_X(p).
$$
\b[Both proposed converses are false.] Take $X=Y=\mathbb R^2$ with the Euclidean metric and $\psi(x)=2x$. A <geodesic> $p+tv$ maps to the <geodesic> $2p+2tv$, and both <Gaussian curvatures> vanish identically. Nevertheless $\psi$ multiplies lengths by two, so is not an <isometry>. This one counterexample answers both unheaded continuation requests.
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