Yes, these are very well-understood! Here's what they are. If $p$ is odd then $\mathbf{Z}_p^\times$ is a direct product of $\mu$, the subgroup of $p-1$th roots of unity, and $1+p\mathbf{Z}_p$, the principal units. A continuous character of the product is a product of continuous characters, so that reduces the first part to the second part. As for the second part, the principal units are topologically generated by $1+p$ so it suffices to say where $1+p$ should go. Note however that $1+p$ can't go to an arbitrary element of $\mathbf{Z}_p^\times$ because you need that if $(1+p)^{n_i}$ tends to 1 in $\mathbf{Z}_p$ then $s^{n_i}$ tends to 1 in $\mathbf{Z}_p$, where $s$ is the image of $1+p$. You can check that, for example, $s=-1$ does not have this property (because the $n_i$ can be even or odd and still tend to zero $p$-adically). But it's also not hard to check that $s$ has this property iff $s$ is a principal unit. I do this in Lemma 1 of my paper "On p-adic families of automorphic forms" here but this is most certainly standard and not due to me.

So in summary, for $p>2$, characters of the principal units biject with $1+p\mathbf{Z}_p$ non-canonically, the dictionary being "image of $1+p$", and characters of the full unit group biject with the product of this and the cyclic group of order $p-1$, that being the characters of $\mu_{p-1}$.

For $p=2$ the two questions are the same, and the same trick, appropriately modified, works. The group $1+4\mathbf{Z}_2$ is procyclic, generated by 5, and its characters biject with the principal units, the dictionary being "image of 5". For the full unit group the characters biject with the principal units product +-1, because $\mathbf{Z}_2^\times$ is just a product $\pm1\times (1+4\mathbf{Z}_2)$.