The number of necklaces that are symmetric after rotation by $i$ is given by \[{\frac{n+m}{\mathrm{ord}(i)} \choose \frac{m}{\mathrm{ord}(i)}}\] if $\mathrm{ord}(i)$ divides $m$ and $0$ otherwise (here $\mathrm{ord}(i)$ is the additive order of $i$ in $\mathbb{Z}_{m+n}$).
To see this, label the cells $0,1,\ldots,n+m-1$. We "populate" the first $\frac{n+m}{\mathrm{ord}(i)}$ cells $0,1,\ldots,\frac{n+m}{\mathrm{ord}(i)}-1$, which determines the rest of the necklace through its symmetry. I.e., if we place a bead in cell $x$, to satisfy the symmetry, we must also place it in the $\mathrm{ord}(i)$ cells $x,x+i,x+2i,\ldots$ (working modulo $n+m$), thereby introducing $\mathrm{ord}(i)$ copies of that symbol.
Hence Burnside's Lemma implies the total number of necklaces is:
\[\frac{1}{m+n} \sum_{{i \in \mathbb{Z}_{m+n}} \atop {\mathrm{ord}(i) \text{ divides } m}} {\frac{n+m}{\mathrm{ord}(i)} \choose \frac{m}{\mathrm{ord}(i)}}.\]
We can make the sum purely numerical by using the identity \[\mathrm{ord}(i)=\frac{n+m}{\mathrm{gcd}(i,n+m)}\] when $i \neq 0$ (and $\mathrm{ord}(0)=1$).