Let's consider $\phi :G\rightarrow G$ an isomorphism such that $\phi (g)=ga$ for some $g\in G$ and $\forall g\in G$, and such that $\phi $ is a cycle including all elements of $G$. Is it then true that $G$ is cyclic?
$\phi$ is a cycle means that it can be expressed in just one cycle in "cycle form."
(I am formulating an alternate definition.)
If $\phi(g)=ga$ is a group homomorphism $G\to G$ then $\phi(e)=a$ implies $a=e$ and $\phi$ is the identity automorphism. If, further, its disjoint cycle representation as a permutation of $G$ is a single cycle containing all of $G$'s elements, then $G$ is the trivial group, since the identity automorphism is a product of $1$-cycles $(g)$ as $g$ runs over all elements of $G$.
However, if you're letting $G$ act on itself by left-multiplication, then $\phi(g)=ga$ will be a $G$-set automorphism for any $a\in G$, and its cycles will be the left cosets of $\langle a\rangle$. Then if $\phi$ is a single cycle, that implies $G=\langle a\rangle$.