I've come across this situation in a number of places but it's most glaring in the lecture notes I'm currently reading. (PCMI lectures on the geometry of outer space).
We have a map from a circle into the $n$ rose $R_n$, $f:S^1 \to R_n$ representing some element of $\pi_1(R_n)$ up to free homotopy (so base points aren't fixed). The notes make the claim that there is a unique immersed representative of $f$ so long as $f$ doesn't represent the trivial conjugacy class.
The main problem I have with this statement is that the $n$ rose isn't a smooth manifold and so speaking of immersions doesn't quite make perfect sense.
I have been toying with the idea of canonically identifying the $n$ rose with the $n$ punctured disk and speaking of smooth representatives in this context and pulling them over. The main barrier I see is that immersed representatives are by no means unique with this representation.
Another approach that you could take is that the $n$ rose without the origin is a smooth manifold and work from there. I think this is the approach that the notes takes, but then you can't speak of immersions onto the origin which is one of the main points. You can represent the nontrivial conjugacy class $a$ with the map $S^1 \mapsto bab^{-1}$ which is an "immersion" aside from the origin. (Here let $a$ and $b$ be one of the non trivial loops in $R_n$). We really want the cyclically reduced representative, but I'm not clearly seeing the correct way of doing this.
Well, I'm not familiar with these notes, but they probably mean the following thing. Note that $R_n \subset \mathbb R^2,$ which is a differentiable manifold. So they are probably constructing an immersion $f:S^1 \to\mathbb R^2,$ with the property that the image of $f$ lies in $R_n$.