TITLE: "Largest" finite-dimensional Lie subgroups of Diff(S^n), are they known?
QUESTION [21 upvotes]: The group $Diff(S^n)$ ($C^\infty$-smooth diffeomorphisms of the $n$-sphere) has many interesting subgroups.  But one question I've never seen explored is what are its "big" finite-dimensional subgroups? 
For example, $Diff(S^n)$ contains a finite-dimensional Lie subgroup of dimension $n+2 \choose 2$, the subgroup of conformal automorphisms of $S^n$.  Similarly it contains a compact Lie subgroup of dimension $n+1 \choose 2$, the isometry group of $S^n$. 
Is it known that:
1) A finite-dimensional Lie subgroup of $Diff(S^n)$ having dimension at least $n+2 \choose 2$ is conjugate to a subgroup of the conformal automorphism group of $S^n$ ? (Answer, no, see Algori's answer below). Modified question: As Algori notes, $GL_{n+1}(\mathbb R) / \mathbb R_{>0}$ acts on $S^n$ and has dimension $n^2+2n$. So is a finite-dimensional Lie subgroup of $Diff(S^n)$ of dimension $n^2+2n$ (or larger) conjugate to a subgroup of this group? 
2) A compact Lie subgroup of $Diff(S^n)$ having dimension at least $n+1 \choose 2$ is conjugate to a subgroup of the isometry group of $S^n$ ? (Answer: Yes, see Torsten Ekedahl's post below)
For example, arbitrary compact subgroups of $Diff(S^n)$ do not have to be conjugate to subgroups of the above two groups -- perhaps the earliest examples of these came from exotic projective and lens spaces.  But I have little sense for how high-dimensional these "exotic" subgroups of $Diff(S^n)$ can be.

REPLY [32 votes]: You can make big Lie groups act effectively on small manifolds by cheating: make the group a product of groups, with each factor acting by compactly supported diffeomorphisms on a different disjoint open subset. So the additive group $\mathbb R^N$ becomes a subgroup of $Diff(S^1)$ by flowing along $N$ commuting vector fields supported in $N$ disjoint arcs.
(added later) A similar cheat: let $a_1,\dots ,a_N$ be linearly independent functions of one variable. Then $a_1(x)\frac{\partial}{\partial y},\dots ,a_N(x)\frac{\partial}{\partial y}$ are independent commuting vector fields in the $x,y$ plane. Modify this example to make it compactly supported if you like.
(added still later) My proposed extension of the first cheat to a semisimple group (comment thread of Torsten's answer) is doomed: Choose a point on the circle and choose an element of SL_2(R) that fixes this point and acts on the tangent space there with eigenvalue c>1. Lifting the group element to the universal covering group in the right way, you get an element g of the latter group that fixes all the points above the given point in the universal covering space of the circle, in each case with eigenvalue c. But now if this line with this action could be embedded in a longer line with trivial action outside then there would be a sequence of fixed points of g with eigenvalue c converging to a fixed point of g with eigenvalue 1, contradiction.