There is a counterexample to this statement. Suppose that G is Z₂ and C is the conjugacy class of the identity element e in G. Then C = {e} and hence |C| = |G|/2. But the element e in C has order 1, not order 2.

So we add the assumption that |G| is not equal to 2 and proceed.

Let x be a member of C. By the orbit equation, |C| = [G: G_x] where G_x = {g in G | gxg^-1 = x} (i.e., G_x is the "stabilizer" or "isotropy subgroup" of x). So |G|/|G_x| = |C| = |G|/2. Thus, |G_x| = 2.

Since x is in G_x, we have x^|G_x| = x² = e. That is, x has order 2. (Note that if x = e, then C = {1} and hence |G| = 2. We have already ruled this out.)