How to convert soluble colorless [HgI₄]²⁻ back to insoluble red HgI₂?
As noted in the comments, to attain mercuric iodide, you need to maintain a balance between the concentration of $\ce{Hg^2+}$ and iodide ion. If there is slight excess of iodide ion, it will form a complex (much like $\ce{KI3}$ from $\ce{KI}$). The complex is popularly known as Nessler's reagent and is prepared pretty much in this way.

Another solution is to decompose the complex to revert back into original products. But I won't recommend this solution. Though, it is expected to give the original reactants, occasionally, decomposition reactions give rise to undesired side products and may release toxic fumes of mercury, iodine, iodide, potassium oxide($\ce{K2O)}$ etc. Do not try to heat the complex. Instead opt for the 1st solution.