= Solution
A <cyclic group> is a <group> generated by a single element: there is $g\in G$ such that every element is $g^n$ for some integer $n$. For a <finite group> of prime order $p$, choose $g\ne e$. Its <cyclic subgroup> $\langle g\rangle$ has order dividing $p$ by <Lagrange's theorem>, and its order is greater than one. Hence $|\langle g\rangle|=p$, and $\langle g\rangle=G$. Thus \b[every <group> of prime order is cyclic].
A <group homomorphism> from $C_{11}=\langle a\rangle$ is determined by $h=\varphi(a)$, since $\varphi(a^j)=h^j$. Its image generator must satisfy $h^{11}=e$, so its <order of a group element> divides $11$. As an element of $C_{14}$ its order also divides $14$. These integers are coprime, so the only possible order is one. Thus $h=e$, and
$$
\boxed{\varphi(a^j)=e\quad\text{for every }j.}
$$
This map is indeed a <group homomorphism>, so \b[there is exactly one], the trivial homomorphism.
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