First divide the table into $11$ equal arcs. Now if the cards $i, j$ be on the points $A, B$ on the table, we define the distance between these two cards as the number of arcs between $A, B$ on the table (the smaller one). For example, the distance between $i, j$ is $5$ in the following figure:

Now, after each step we sum the distances between every two cards with consecutive numbers. Easily it can be proved that after each step this sum either remains invariant or varies by $2$. Because if the place of card $i$ is between $i-1$ and $i+1$, this sum doesn't change, otherwise it changes by $2$. So the parity of this sum remains invariant. At first it is odd ($= 11$), and if they are all in one place this sum would be even, so it is impossible.