Let $H$ be cyclic group of order 7 and let $K$ be cyclic of order 3.
(a) Determine all homomorphisms $\phi: K\rightarrow Aut(H)$.
My attempt: Since $K$ is cyclic, any one of the two generator of $K$ will be mapped as the generator of $Im(K)$ in $Aut(H)$. Since "if $f$ is a homomorphism, then $|f(x)|||x|$.", so the $|Im(K)|||K|$ thus $|Im(K)|$ could be 1 or 3. Since $H$ is also cyclic, $Aut(H)\cong\mathbb{Z}_7^X$, so the size, by Euler's function, of $Aut(H)$ is 6. Then I'm stuck...
(b) For each homomorphism describe the corresponding semi direct product $H\rtimes_\phi K$.
My attempt: Once I have all the homomorphism from (a), I'll have the operation of $(h,k)(h',k')$. I'm not sure what else I'm suppose to say for the "description."
(c) Determine which semi direct products from (b) are isomorphic to each other.
I'm not sure how to do this...
First of all: $\text{Aut}(\Bbb Z_7) \cong \Bbb Z_6$.
So you're looking for homomorphisms between two cyclic groups, which simplfies things.
Secondly, it suffices to consider $\phi(1)$ (or more generally, any generator of $K$). The image of such a generator can only be of order $1$, or order $3$.
There are thus three possible homomorphisms, resulting in two distinct isomorphism classes of semi-direct products.
That should get you started.
By "description", what is probably meant is describing them in terms of groups you already know, or by generators and relations. The non-abelian group may prove a bit harder than the abelian one.