Classifying Groups of Order 42 with an element of Order 6

Question

Let $G$ be a group of Order 42 which has an element of order 6. I want to classify all groups like this up to isomorphism.
Let $g\in G$ be the element of order 6. Then $C_6\cong\langle g\rangle\leq G$. As $42=6\cdot 7$ and $6<7$ there is only one 7-Sylow Subgroup, which thus is normal and ismorphic to $C_7$. Let $P$ be this group.
As elements in $P$ have order 7 and elements in $\langle g\rangle$ have order which divides 6, we know that $P\cap \langle g\rangle=\{1\}$.
Now as P is normal, the Frobenius product of P and $\langle g\rangle$ is a subgroup of Order $$\frac{|P||\langle g\rangle|}{|P\cap \langle g\rangle}=42$$ So we know the Frobenius Product is the whole Group thus the Group is its inner semidirect bewteen P and $\langle g \rangle$.
We know now, that all Groups like this are ismorphic to a group of the form $C_7\rtimes_\varphi C_6$ for an homorphism $\varphi$ bewteen $C_6$ and $Aut(C_7)$ and all Groups that are of this form have $C_6$ as a subgroup and are of order 42.
So we need to find all homorphisms between $C_6$ and $Aut(C_7)$. This is where I'm now stuck a litte bit. How would I go about classifying all those homomorphisms?