Note that $O(1)$ is the group $\{\pm 1\}$. Any homomorphism into $\mathbb C^*$ sends $1$ to $1$ and $-1$ to some square root of $1$, that is to either $1$ or $-1$. Thus there are two maps in $Hom(O(1),\mathbb C^*)$, the constant map $1$ and the identity, so $Hom(O(1),\mathbb C^*)\cong \mathbb Z/2\mathbb Z\cong O(1)$.
On the other hand, any homomorphism from $\mathbb C^*$ to $O(1)$ has kernel of index $1$ or $2$. No subgroup of $\mathbb C^*$ has index $2$, or any proper subgroup of finite index. To see this, note that $\mathbb C^*$ can be factored into the circle group and the real numbers under addition (by mapping $re^{i\theta}$ to $(e^{i\theta},\log r)$). A finite index subgroup of $\mathbb R$ would give us a decomposition $\mathbb R=H\cup (a_1+H)\cup \cdots \cup (a_n+H)$. Similarly, restricted to the rational numbers it gives us the rationals as a union of disjoint cosets. We can restrict our attention to those $a_i$ that are rational. We have some subfield $K$ which contains none of the $a_i$, and the denominators in $K$ are coprime to all $a_i$, thus $K\subseteq H$ as it can contain no element of the other cosets. We can then take realize $H$ as a vector subspace of $\mathbb R$ over $K$, which must correspond to a proper subset of some basis for $\mathbb R$. Thus the quotient of $\mathbb R$ by $H$ as a vector space contains at least one copy of $K$, which is infinite, contradicting the fact that $H$ has finite index. Thus any finite index subgroup would have to correspond to a finite index subgroup of the circle group, but this is isomorphic to $\mathbb R/\mathbb Z$ and the same approach works here (I think). Thus the kernel has index $1$ so is all of $\mathbb C^*$, hence the only homomorphism from $\mathbb C^*$ to $\mathbb C^*$ is the trivial one, and so $Hom(\mathbb C^*,O(1))$ is the trivial group.
As a corollary, the two groups are not isomorphic.