= Solution
Put $q=p^n$ and choose a generator $h$ of $H$. In characteristic $p$, $(h-1)^q=h^q-1=0$, so
$$
kH\cong k[t]/(t^q),\qquad t=h-1.
$$
A <module> is a vector space with a <nilpotent operator> $t$ whose nilpotence index is at most $q$. The nilpotent <Jordan normal form>, valid over every <field>, splits it into Jordan blocks. One block of size $i$ is
$$
\boxed{V_i=k[t]/(t^i),\qquad1\le i\le q.}
$$
It is indecomposable: its <endomorphism ring> is $k[t]/(t^i)$, a <local ring> with no nontrivial <idempotents>. Distinct $i$ have distinct dimensions. Conversely a <direct sum> of at least two Jordan blocks is decomposable. Thus there are exactly $q$ indecomposable isomorphism classes.
A homomorphism $V_i\to V_j$ is determined by the image of $1$, which can be any element of $V_j$ annihilated by $t^i$. If $i\ge j$ this is all of $V_j$, of dimension $j$; if $i<j$ it is the span of $t^{j-i},\ldots,t^{j-1}$, of dimension $i$. Hence
$$
\boxed{\dim_k\operatorname{Hom}_{kH}(V_i,V_j)=\min(i,j).}
$$
This realizes the <indecomposable modules for a cyclic p-group>.
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