= Solution
{c}
Write $C_p=\langle g\rangle$. A <group homomorphism> from this <cyclic group> is determined by the image of $g$, because $\phi(g^j)=\phi(g)^j$. Each choice $\phi(g)=g^a$, with $a=0,\ldots,p-1$, gives a well-defined <group homomorphism>. It is an <group automorphism> exactly when $g^a$ generates $C_p$.
For a <prime number> $p$, every $a=1,\ldots,p-1$ is <coprime> to $p$, so the powers of $g^a$ run through all of $C_p$. The choice $a=0$ gives the trivial <group homomorphism> and fails to be bijective. Consequently
$$
\boxed{|\operatorname{Aut}(C_p)|=p-1.}
$$
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