Dylan's right that diffeomorphism groups are usually infinite dimensional. There's a theory of infinite dimensional Lie groups, I've heard, but I think it's a lot harder to work out things like convergence. Worth noting, though, is that in this analogy, if $Diff(M)$ is a Lie group then its Lie algebra should be $Vect(M)$, the (infinite-dimensional) vector space of smooth vector fields on $M$. The Lie algebra bracket is just the usual vector field bracket (possibly up to a sign? but if so, then it's only because of stupid clashing conventions). Heuristically, to look for finite dimensional Lie groups we're going to want to put some more structure on $M$ and require that our diffeomorphisms preserve this. If you're looking at things that aren't so homogeneous you get into the territory of $G$-bundles, which I think you should definitely look into if you don't already know about -- you don't need to know too much Lie theory to get the basics, and it's really cool (I think you'll like it too). I don't have a reference offhand (please, anyone, post as a comment if you do) but I took an excellent class on them from Robert Bryant last year and a friend of mine is working on TeXing up the notes, so hopefully those will be available (at least in part) before too long.
By the way, $Aut(M)/$isotopy is called the mapping class group. This can be written $MCG(M)=Aut(M)/Aut_0(M)$, where $Aut_0(M)$ is the connected component of the identity map, which if you think about it is all isotopies are anyways. This has been extensively studied. I don't remember much from a class I took with Jeff Brock a few years ago, but checking wikipedia reminds me that e.g. $MCG(\mathbb{T}^n)\simeq GL(n,\mathbb{Z})$. It also says that the MCGs of surfaces have been extensively studied. In any case, you can see that this is just the group $\pi_0(Aut(M))$, so assuming you understand $MCG(M)$ (!) then you only need to figure out $Aut_0(M)$ (!) to have the whole picture. Which brings us back to the original question.
I don't know much, but like I said it's definitely easier (and finite-dimensionaler) if we restrict ourselves to manifolds with structure. We can start by saying something simple: $Isom(S^n)=O(n+1)$. One potentially fruitful way to think about this is to just start with a (matrix) group and find an invariant subspace of the (affine) space on which it acts, hopefully freely -- very often this should be a manifold, I think. Otherwise things will probably be significantly more difficult. If you put the flat metric on a torus and assume its fundamental domain is a square, then you should be able to look at isometries by what they do to your fundamental square sitting on a latticed plane. To represent an isometry it'll need to be an isometry of the square, and to be a continuous map you'll need the four corners to be mapped to preimages of the same point under $\mathbb{R}^2\rightarrow \mathbb{T}^2$. I'm pretty sure this ends up meaning $Isom_0(\mathbb{T}^2)\cong \mathbb{T}^2$. But maybe that's not so surprising, since it's a group itself. I guess all it's really telling you is that there are no isometries other than the "multiply-by-$g$" map for $g\in \mathbb{T}^2$. The group $Conf(\mathbb{T}^2)$ should be pretty different, but I think you can't continuously dilate because that would eventually change the degree of your map (which is a discrete invariant), so a similar analysis should go through but you won't get any interesting Lie-type behavior. More interesting might be to try this all with the fundamental domain for an $n$-holed torus ($n\geq 2$) sitting on a latticed hyperbolic plane. But it's late and I'm tired, so I'm not going to. You should though! If you do, let me know what you come up with.