On solvable instances (not all instances are solvable; see narek Bojikian's comment on my other answer for a simple necessary and sufficient condition), this is NP-hard by reduction from Directed Hamiltonian Path. The basic idea is to add enough isolated vertices to the input digraph to force the only possible solution to be a single cycle, and then see whether it is possible to accomplish this with few enough moves that the unmoved edges must form a DHP in the original digraph.
Given an instance of Directed Hamiltonian Path consisting of a digraph $(V, E)$ with $n=|V|, m=|E|$, add $m-n$ isolated vertices. Observe that a single SCC can be formed by rearranging the $m$ edges into an arbitrary cycle that touches each of the $m$ vertices that now exist. This can be done trivially if we are prepared to move all $m$ edges, but if we limit ourselves to moving at most $m-n+1$ edges, then this is possible only if $m-n+1$ edges are moved from somewhere to form a path that connects some vertex $u$ of the original digraph to some new isolated vertex, visits each new isolated vertex in some order, and finally returns to some other vertex $v$ of the original digraph. (We have $u \ne v$ since $u=v$ would leave just $n-1$ edges available to visit all $n$ vertices of the original digraph, which is not enough.) In this case, the unmoved $n-1$ edges that remain necessarily form a path from $v$ to $u$ through the $n$ vertices of the original digraph, and thus constitute a DHP through it. This shows that a YES answer to the constructed instance implies a YES answer to the original problem.
To see that a YES answer to the original DHP problem implies a YES answer to the constructed instance, pick up the $m-n+1$ edges that are not part of the DHP, and construct a path as described above, forming a single cycle comprising all $m$ vertices using just $m-n+1$ moves.