At poles, a stretched wire of a given length vibrates in unison with a tuning fork. At the equator, for same setting, to produce resonance with same fork, the vibrating length of wire

The frequency of a vibrating stretched wire is given by $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$, where $L$ is the wire length, $T$ is the tension, and $\mu$ is the linear mass density. For resonance with the same tuning fork, $f$ remains constant, and $\mu$ is unchanged.

The tension $T$ equals the weight of the hanging mass, so $T = mg$. At the poles, $g$ is unchanged, but at the equator, the effective gravity is reduced to $g - R\omega^2$, where $R$ is Earth's radius and $\omega$ is its angular velocity. Consequently, $T_{\text{equator}} = m(g - R\omega^2) < T_{\text{pole}} = mg$.

Since $f \propto \frac{1}{L}\sqrt{T}$ and $f$ is constant, $L \propto \sqrt{T}$. With $T_{\text{equator}} < T_{\text{pole}}$, it follows that $\sqrt{T_{\text{equator}}} < \sqrt{T_{\text{pole}}}$, so $L_{\text{equator}} < L_{\text{pole}}$ to maintain the same frequency.

Thus, the vibrating length must be decreased, and the correct choice is $\boxed{A}$.