Step 1: Verify (I) by exhibiting the isomorphism explicitly.
Define $\phi:\mathbb{Z}_{15}\to \mathbb{Z}_3\times\mathbb{Z}_5$ by $\phi(x) = (x \bmod 3,\ x \bmod 5)$. This is a homomorphism since reduction mod $3$ and mod $5$ are both homomorphisms. Its kernel consists of all $x$ with $3\mid x$ and $5\mid x$, i.e. $15\mid x$, so $\ker\phi = \{0\}$, meaning $\phi$ is injective. Since $|\mathbb{Z}_{15}| = |\mathbb{Z}_3\times\mathbb{Z}_5| = 15$, injectivity forces $\phi$ to be a bijection, hence an isomorphism. So (I) is true.
Step 2: Verify (II) by computing conjugation directly.
$A_3$ is abelian (cyclic, generated by $(123)$), so for every $g,h\in A_3$, $ghg^{-1}=h$. So every inner automorphism $\iota_g:h\mapsto ghg^{-1}$ of $A_3$ is the identity map. So\[\text{Inn}(A_3) = \{\iota_g : g \in A_3\} = \{\text{id}\}\]which has one element, while $A_3$ has three elements. A group of order $1$ cannot be isomorphic to a group of order $3$, so $\text{Inn}(A_3) \not\cong A_3$, and (II) is false.
Step 3: Conclusion.
(I) is correct and (II) is the incorrect one.\[\boxed{\text{Only (II)}}\]