First, it must be clear that in any group of order $\,27\,$ , a subgroup of order $\,9\,$ is normal (why? what is this subgroup's index?)
Second, as was already pointed out above, you already know that
$G=C_3\ltimes H\,\,\,,\,\,C_3:=\langle x\;\;;\;\;x^3=1\rangle$
We now have two possible cases:
$(1)\;\;H= C_9\;\;\Longrightarrow\,\operatorname{Aut}H\cong C_6\;\;(\text{Hint:}\,\,\phi(3^2)=3\cdot 2=6)$
The only non-trivial homomorphisms $\,C_3\to C_6=\langle y\rangle\,$ are $\,x\to y^2\,\,,\,\,x\to y^4\,$ , which give us non-trivial groups of order $\,27\,$ . These two though are isomorphic, as you can read in page $\,49-50\,$ in the PDF here
$(2)\;\;H=C_3\times C_3\;\;\Longrightarrow\,\operatorname{Aut}(H)\cong\,C_2\times C_2$
The only homomorphism possible here is the trivial one (why?), giving us a direct product and thus an abelian group.