Equivariant map in compact homogeneous space is a diffeomorphism I don't see why an equivariant $G$-map $f: M \rightarrow M$, where $M$ is a compact homogeneous space, is necessarily a difeomorphism. Any idea?
 A: I just wanted to complement freakish's (perfectly fine) answer with details about smoothness.  To that end, recall that any homogeneous space $M$ is equivariantly diffeomorphic to a coset space $M\cong G/H$, where the $G$ action on $G/H$ is just left multiplication.  I'll use the notation $N_G(H)$ for the normalizer of $H$ in $G$.
Lemma:  Suppose $f:G/H\rightarrow G/H$ is equivariant.  Then there is a $g\in N_G(H)$ for which $f(kH) = kgH$ for all $kH\in G/H$.
Proof:  Choose $g\in G$ with $f(eH) = gH$ (with $e\in G$ the identity).  Then, for any coset $kH$, we have $f(kH) = f(keH) = kf(eH) = kgH$.
To see $g\in N_G(H)$, note that for any $h\in H$, $f(kH) = f(khH)$, which implies that $kg = khg h'$ for some $h'\in H$.  Canceling $k$ and rearranging gives $g^{-1} h^{-1} g \in H$, which then easily implies $g\in N_G(H)$.  $\square$
Now, if $\pi:G\rightarrow G/H$ is the obvious projection and $R_g$ is right multiplication by $g$, it follows that $f\circ \pi = \pi \circ R_g$.  The map $\pi\circ R_g$ is obviously smooth, so $f\circ \pi $.  Because $\pi$ is a submersion, smoothness of $f\circ \pi$ implies smoothness of $f$.  (See this MSE question for a proof of this last statement.)
Finally, to see $f$ is a diffeomorphism, simply note that if $f(kH) = kgH$, then $f^{-1}(kH) = kg^{-1} H$, so it's smooth for the same reason $f$ is.
A: Partial answer:
So $f(gx)=gf(x)$ for any $g\in G$, $x\in M$. Since $M$ is homogenous this means that $f$ is uniquely determined by a single value on a single element $x_0\in M$, say $y_0=f(x_0)$. Therefore we can construct the inverse $h:M\to M$ by $h(y_0)=x_0$ which uniquely extends to entire $M$ via $h(gy_0):=gh(y_0)$.
And so $f$ is a continuous bijection. Which means it is a homeomorphism since $M$ is compact.
Now you also ask about diffeomorphism so I assume that $M$ is additionally a smooth manifold. Typically smooth homeomorphisms need not be diffeomorphisms. The $x\mapsto x^3$ map is a smooth map that is not a diffeomorphism (the derivative at $0$ is $0$). I think we can turn that example into compact case by considering $[-1,1]$ interval and glueing ends into $S^1$. But this example does not include the action of $G$ on $M$, so I might actually be wrong here.
