Let $X$ be a space equipped with an action of a compact Lie group $G$. Recall that such a space is said to be $G$-contractible if the identity map of $X$ is $G$-homotopic (i.e., homotopic through $G$-maps) to a map with values in a single orbit.
In general, a $G$-contractible space $X$ does not need to be contractible (e.g. any Lie group $G$ considered as a $G$-space with the action given by the group multiplication is $G$-contractible). However, if $X$ has a fixed point, then $G$-contractibility implies contractibility.
My question is: does contractibility imply $G$-contractiblity? I suspect the answer is "no" in general, so a complementary question would be: under what additional assumptions this is true?
This is an interesting question. Here's a counterexample that should work for any compact Lie group: the total space $EG$ of the classifying bundle $G \to EG \to BG$. This is a contractible space with a free $G$-action. For $G$ nontrivial, $BG$ won't be contractible, so $EG$ can't be $G$-contractible; if it were, taking quotients by $G$ of the given homotopy would give a homotopy from the identity map of $BG$ to a point.
One thing to draw from this is that an obvious necessary condition for $X$ to be $G$-contractible is for $X/G$ to be contractible. This isn't sufficient either, though: for example, let $X$ be the circle with $G = \mathbb{Z}/2$ acting by reflecting about the equator. Then $X$ has a fixed point and $X/G$ is contractible, but $X$ clearly isn't $G$-contractible (you'd have to deformation retract the circle to a point, or worse yet, two points).
The right condition, as it turns out, is what's called the 'equivariant Whitehead theorem': if $X \to Y$ is a $G$-map (of $G$-CW-complexes) and the map on fixed-point subspaces $X^H \to Y^H$ is a homotopy equivalence for every closed subgroup $H \le G$, then $X \to Y$ has a $G$-equivariant homotopy inverse (and conversely). In your case, $Y$ is a point, so you want $X^H$ to be contractible for each subgroup $H$. Notice how this fails in the two examples above: when $X = EG$, $G$ acts freely, so each fixed-point set is empty, not contractible; and when $X = S^1$ with the $\mathbb{Z}/2$-action, the fixed point set has two points rather than one.
EDIT: Here's a related MO question. The last paragraph of Tom's answer beefs up my $S^1$ example in the following awesome way. There are spaces $Z$ with a nontrivial fundamental group but no homology, like the classifying space of any perfect group. Then the suspension $\Sigma Z$ is simply connected and has no homology, so it's contractible. If $G = \mathbb{Z}/2$ acts on this by flipping it about the equatorial $Z$, then the orbit space is $\Sigma Z/G = CZ$, which is also contractible, but by the equivariant Whitehead theorem, $\Sigma Z$ isn't $G$-contractible since $Z = (\Sigma Z)^G$ isn't contractible. This is an example of a $G$-space $X$ such that $X/H$ is contractible for every $H \le G$, but $X$ isn't contractible, so the analogue of the theorem with 'fixed points' replaced by 'orbits' is false. One wonders how many more examples like this there are out there.