Answer: $\frac{9n^2-7n}{2}$ moves.

Below, $EX$ denotes the expected value of the random variable $X$.

1. First, for any two points $A$ and $B$, let $N_{AB}$ be the number of steps required to make $A$'s laser point at $B$. Suppose we choose distinct $A$ and $B$ uniformly at random from all the points, then $EN_{AB} = \frac{3n-2}{2}$ (in fact, it is easy to see that $N_{AB}$ is uniformly distributed on $\{0,1,\dots,3n-2\}$.)

2. For any three points $A$, $B$ and $C$, let $N_{ABC}$ be the number of steps required to rotate the lasers on the three points so that $ABC$ becomes a triangle. Suppose we choose distinct triples $(A,B,C)$ uniformly at random from all the points, then by additivity of expectation, we know $EN_{ABC} = EN_{AB} + EN_{BC} + EN_{CA} = \frac{3(3n-2)}{2}$.

3. Denote these $3n$ points by $A_1$ through $A_{3n}$. We use $T = (T_1,T_2,\dots,T_n)$ to denote a way of partitioning all the points into $n$ triangles, where $T_i = (A_{i1},A_{i2},A_{i3})$ is an ordered triple. Let $N_T$ denote the number of steps required to turn the given configuration into $T$. Suppose we choose $T$ uniformly at random among all possible partitions, then as in point 2, by additivity, $EN_T = \frac{3n(3n-2)}{2}$.

4. We use $S = (S_1,S_2,\dots,S_n)$ to denote a way of partitioning all the points into $n$ unordered sets, each of three points. For each $T$, let $S(T)$ be the corresponding unordered grouping of $T$. Note that the random variable in the third point can be viewed as first choosing an $S$ uniformly at random among all possible $S$, and then considering all permutations of $S_i$ uniformly at random. Based on $EN_T = \frac{3n(3n-2)}{2}$, this means there exists an $S^*$ such that $E[N_T|S(T) = S^*] \le \frac{3n(3n-2)}{2}$.

5. Now, note that all $T$ satisfying $S(T) = S^*$ have the same grouping into triples of points, differing only in the clockwise or counterclockwise orientation of some of the triangles. Consider the following proposition:

Proposition 1. For a triangle, the number of steps required to turn it clockwise differs from the number of steps required to turn it counterclockwise.

If Proposition 1 holds, then it means $\{N_T|S(T) = S^*\}$ contains at least $n+1$ distinct values, so among all $T$ satisfying $S(T) = S^*$ there must exist a $T^*$ such that $N_{T^*} \le E[N_T|S(T) = S^*] - \frac{n}{2} \le \frac{3n(3n-2)}{2} - \frac{n}{2} = \frac{9n^2-7n}{2}$, thereby obtaining the claimed lower bound.

6. To prove Proposition 1, consider an unordered triangle $ABC$, and suppose that the number of steps required to turn it into a clockwise triangle is the same as the number of steps required to turn it into a counterclockwise triangle. Without loss of generality, assume $ABC$ is a clockwise triangle; furthermore, let $n_1$ be the number of steps required to rotate $BC$ clockwise to $BA$, $n_2$ be the number of steps required to rotate $CA$ clockwise to $CB$, and $n_3$ be the number of steps required to rotate $AB$ counterclockwise to $AC$, then we have $n_1 + n_2 = n_3$. However, note that rotating $BC$ clockwise to $BA$, together with rotating $CA$ clockwise to $CB$, will pass through every point on the arc $BAC$ exactly once, but rotating $AB$ counterclockwise to $AC$ will pass through every point on the arc $BAC$ "except $A$", which means the two operation counts differ by one, so the original assumption cannot hold.

7. Finally, we prove that this lower bound is optimal. Consider a regular $3n$-gon, whose vertices are labeled clockwise in order as $A_1, A_2, \dots, A_{3n}$, and consider the situation where robot $A_1$ points at $A_2$, while all other robots point at $A_1$. Consider the case where, in the end, $a$ counterclockwise triangles are turned out; then one can compute that $\frac{3n(3n-1)}{2} - 2n + a$ operations are required, so when $a = 0$ we get the minimum value $\frac{3n(3n-1)}{2} - 2n = \frac{9n^2-7n}{2}$.