Order of $\beta^{11}$ is 5.
hence, $\frac{n}{(n, 11)}=5$.
If $11|n \implies n=55$.So $\beta$ is a combination of 2 cycles 5 and 11.
Let $\beta =(a_1, a_2,a_3,a_4,a_5)(a_6,a_7,a_8,a_9,a_{10},a_{11},a_{12},a_{13},a_{14},a_{15},a_{16})$.Now the first 5 cycle will be equal to $(12893)$ and for the remaining elements I have a choice of $11$ elements out of $15$.So predicting the element is not possible in this case
If $n=5$,then $\beta$ is a 5 cycle so $\beta^{10} \beta =(12893) \implies \beta=(12893)$
This has been my attempt. Is this OK?