It seems to me that the structure of ${\rm Aut}^{m}(C_{n})$ also depends heavily on the prime factorization of $n$, and I don't really see any reason to expect the answer to Q1 to be any more tractable than determining the structure of ${\rm Aut}(C_{n})$.

For example (just to illustrate) , if we choose a prime $p$, greater than $3$, and then we take a pair of primes $q_{1}$ and $q_{2}$ so that each $q_{i}-1$ is divisible by $p$, but not by $p^{2}$, and we set $n = q_{1}q_{2}$, then ${\rm Aut}^{2}(C_{n})$ has a direct factor isomorphic to ${\rm GL}(2,p)$, and is not solvable. Furthermore the involvement of ${\rm GL}(2,p)$ in ${\rm Aut}^{m}(C_{n})$ persists for $m >2$.

It seems to me that the structure of ${\rm Aut}^{m}(C_{n})$ also depends heavily on the prime factorization of $n$, and I don't really see any reason to expect the answer to Q1 to be any more tractable than determining the structure of ${\rm Aut}(C_{n})$.

For example (just to illustrate) , if we choose a prime $p$, greater than $3$, and then we take a pair of primes $q_{1}$ and $q_{2}$ so that each $q_{i}-1$ is divisible by $p$, but not by $p^{2}$, and we set $n = q_{1}q_{2}$, then ${\rm Aut}^{2}(C_{n})$ has a direct factor isomorphic to ${\rm GL}(2,p)$, and is not solvable. Furthermore the involvement of ${\rm PSL}(2,p)$ in ${\rm Aut}^{m}(C_{n})$ persists for $m >2$.