That question does not have a unique answer. For instance one possibility is to take a beaker as your cylinder, fill it halfway with clay as your shape, scrape a thin layer off the top, form it into a little straight toothpick of clay the height of the beaker, and then insert the toothpick into the very center of the clay in the beaker so that it reaches to the top. This shape is rotationally symmetric, it has a volume of half the beaker, and the beaker is the smallest cylinder that encloses it. However you could do this toothpick idea to just about any shape that you could throw out of that same volume of clay on a potter's wheel, and you might not even need to.
However one of the easiest algebraic ways to do it is to have the radius go like the square root of the distance from the top of the beaker, then the cross-sectional area goes linearly with that distance and when you integrate the integration gives you a clean factor of ½ rather than one-third.
Technically correct, but I think it can be assumed from the question that it's referring specifically to a conical frustum. Of course, the answer is then that the shape of such an object would be, well, a conical frustum, which renders the question tautological.
I believe that the intent of the question is something like “What are the dimensions of a conical frustum having 1/2 the volume of a cylinder, with the same base and height?”, or, put another way, what would the ratio of the diameter of the top of such a frustum be to its base? The answer being somewhere between 0 (as it would be for a cone) and 1 (for the cylinder). I'm sure there's a formula to calculate this value for a given volume, but I don't really care to figure it out at the moment. Although, it might be interesting to see if such a formula scales to values for the relative volume above 1, creating an “inverted” conical frustum…
Perhaps he's also asking for some interpolating curve, see my sibling comment. You can step through the numerical program to see how the shape converges:
