Random Shuffle of Groups Let's suppose that we have 54 peoples and we arrange them into 9 groups of equal size, so this means that each group will have 6 persons in it. I want to find a procedure, such that the groups are shuffled deterministic in order that people will not meet several times in a group. I couldn't come of with an idea so far.
Since each person will meet then 5 new people in each group, this means that we can shuffle the groups up to 10 times. So I will decrease the complexity of this question to come up with a mix such that we will build 10 times new groups. Adding a random element in the shuffling would be an extra!
 A: Note that $54 = 2 \times 3^3$. This can be visualized as stacking two Rubik's cubes on top of each other. Just imagine that the cubes can be rearranged anywhere within a layer including the central position.

Let us say the top layer of cubes represents 9 individuals who are the group leaders and they remain at their assigned positions throughout the shuffles. Let other 5 layers of cubes represent other members who can move around to other groups under other leaders. Note that there is no need for them to move across layers, because doing so will pair them up with the same leader (who is situated vertically on the top layer). It is easy to see this holds for all layers. A group at any time instant can be visualized as a single column of vertically stacked cubes in this formation.
We can assign positions $0$-$7$ for each of the boundary (non-central) cubes in a layer and the position $9$ to the central cube in the layer. In each shuffle, the cubes are rearranged by incrementing their position modulo $9$ by a certain number $c_k$ using the relation:
$$pos_k(n) = pos_k(n-1) + c_k \mod 9$$
where, $pos_k(n)$ is the new position for a cell previously in position $n-1$ in a layer $k$.
$c_k \in \{ 1, 2, 5, 7\}$ only because if we choose $3, 6, 9$, there will be shorter periods modulo $9$ and if we choose $4, 8$, there will be a duplication with $2$. Any number greater than $9$ reduces to one of these cases modulo $9$.
But we have 5 layers under the top layer and we cannot choose a value for $c_k$ for one of the layers. Therefore, a duplicate pairing will occur in the first shuffle itself. Moving the leaders to different positions doesn't solve this problem.
So, to answer the question: 10 shuffles are not possible with 54 people without duplicate pairings. After the initial arrangement, not even 1 shuffle is possible.
On the question of replacement: Replacing an entire layer of 9 people in each shuffle with new people will allow 4 layers to take on different positions with the last layer filled with all new people in each shuffle that permits grouping without any duplicate pairing.
