To prove that due to the unknown colors and arrangement of the initial 4 chess pieces, the situation is relatively complex according to the problem statement. Below, we construct a valuation model that reflects the requirements of the problem, which can simplify the problem and thus provide a proof.

Let the initial 4 chess pieces be $x_{i} (i=1,2,3,4)$, and assign values to the chess pieces.
Let
$$x_{i}=\left\{\begin{array}{l} 1, \text { if } x_{i} \text { is a black piece, } \\ -1, \text { if } x_{i} \text { is a white piece. } \end{array}(i=1,2,3,4)\right.$$

And define $x_{i} x_{i+1}=\left\{\begin{array}{c}1 \text {, if } x_{i} \text { and } x_{i+1} \text { are the same color, } \\ -1, \text { if } x_{i} \text { and } x_{i+1} \text { are different colors, }\end{array}\right.$ and $x_{i}^{2}=1$.

After the first operation, the 4 chess pieces can be represented as
$$\left(x_{1} x_{2}\right),\left(x_{2} x_{3}\right),\left(x_{3} x_{4}\right),\left(x_{4} x_{1}\right).$$

After the second operation, the 4 chess pieces can be represented as
$$\left(x_{1} x_{2}\right)\left(x_{2} x_{3}\right),\left(x_{2} x_{3}\right)\left(x_{3} x_{4}\right),\left(x_{3} x_{4}\right)\left(x_{4} x_{1}\right),\left(x_{4} x_{1}\right)\left(x_{1} x_{2}\right),$$
which simplify to $\left(x_{1} x_{3}\right),\left(x_{2} x_{4}\right),\left(x_{3} x_{1}\right),\left(x_{4} x_{2}\right)$.

After the third operation, the 4 chess pieces can be represented as
$$\left(x_{1} x_{3}\right)\left(x_{2} x_{4}\right),\left(x_{2} x_{4}\right)\left(x_{3} x_{1}\right),\left(x_{3} x_{1}\right)\left(x_{4} x_{2}\right),\left(x_{4} x_{2}\right)\left(x_{1} x_{3}\right),$$
which simplify to $\left(x_{1} x_{2} x_{3} x_{4}\right)$.

After the fourth operation, the 4 chess pieces are all $\left(x_{1} x_{2} x_{3} x_{4}\right)^{2}$, so the values of these 4 chess pieces are all 1.
This indicates: after only 4 operations, all the pieces on the circle are black.