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The hydrocarbon $\mathrm{HC}\equiv \mathrm{C}-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}\equiv \mathrm{CH}$. Its triple-bond fragments are linear, but the entire carbon skeleton need not be. The extended anti conformation has an inversion centre at the central bond midpoint and belongs to the C2h molecular point group.

For hexa-1,5-diyne in the extended anti conformation drawn, inversion at the midpoint of the central carbon–carbon bond exchanges the two identical ends. A twofold axis through that midpoint perpendicular to the carbon skeleton plane, together with the mirror plane of that skeleton, completes the C2h molecular point group. The methylene hydrogens occur in the corresponding mirror-related pairs. Thus the answer is $C_{2h}$. The whole molecule is not linear simply because its terminal triple-bond fragments are linear; the conformation is essential to this assignment.