I just started learing about discrete valuation rings, so I don't know a lot of examples and I can't find any counterexamples.
So consider a DVR $(R,v)$ with valuation group $\Gamma_v \neq \{0\}$. A uniformizer of $R$ is an element $u \in R$ such that $v(u) > 0$ generates $\Gamma_v \simeq \mathbb{Z}$. My immediate flow of thought is the following:
We have two generators of $\mathbb{Z}$, namely $1$ and $-1$. Under the isomorphism for $\Gamma_v$, those elements map to two possible generators of $\Gamma_v$. Let's say that $\langle \gamma \rangle = \Gamma_v$, where $\gamma > 0$. By definition, there exist elements $u \in R$, such that $v(u) = \gamma$. Now I wonder if this $u$ is unique, or if there are any conditions in $R$ such that this $u$ is unique.
A general example of DVR's that I know of, are a construction with UFD's. So if $R$ is a UFD, take an irreducible element $p \in R$. Now let $v_p$ denote the $p$-adic valuation. The DVR corresponding to this construction is just the localization $R_{pR}$ of $R$ at $pR$. Moreover, $p$ itself is a uniformizer of $(R_{pR},v_p)$. Since $R$ is a UFD and $p$ is irreducible, it should be unique right?
Also, DVR's are local (as far as I know), which gives us uniqueness of a maximal ideal. I guess if uniqueness of uniformizers does hold, it must have something to do with uniqueness of the maximal ideal. My guess would be, that if the maximal ideal $m$ is principal, then the generator would be this generator of $m$?
Of course it may be possible to come up with very complicated examples, which might have multiple uniformizers. But are there any (more or less) simple counterexamples? And if so, are there any conditions to ensure uniqueness of uniformizers?
A uniformizer is indeed the generator of the (unique) maximal ideal $m$, which is just the subset of elements $r\in R$ with $v(r)>0$. It is hence not unique, since for any unit $u\in R$ if $a$ generates $m$, then $ua$ also does (note the change of notation from your post). But that's it!
Assume $a,b\in m$ are two generators, i.e. $m=(a)=(b)$. Then $b=ua$ for some $u\in R$, and then you immediately find that $v(u)=0$, so $u$ must be invertible.
So the uniformizer is unique up to multiplication by a unit.