Orientation reversing diffeomorphism Why each sphere admits an orientation reversing diffeomorphism onto itself? (For even dimensional ones can we take conjugation map?) And why complex projective spaces do not admit? Is there a geometric way to see this without using characteristic classes?  
 A: Complex spaces of the form $\mathbb{C}P^{2n+1}$ do admit orientation reversing diffeomorphisms.  In fact, the map sending $[z_0:z_1:...:z_{2n}:z_{2n+1}]$ to $[-\overline{z_1}:\overline{z_0}:...:-\overline{z_{2n+1}}:\overline{z_{2n}}]$ is orientation reversing (and gives a free action of $\mathbb{Z}/2\mathbb{Z}$ on $\mathbb{C}P^{2n+1}$).
One can check that when $n=0$, this map is the usual antipodal map of $\mathbb{C}P^1 = S^2$.
Also, Jim's idea works for showing that the manifolds $\mathbb{H}P^{2n}$ and $\mathbb{O}P^2$ do not admit orientation reversing diffeomorphisms.  But more is true.  The manifolds $\mathbb{H}P^{2n+1}$ also do not admit orientation reversing diffeomorphisms (unless $n=0$).  (The map I wrote above isn't well defined when considering $\mathbb{H}P^n$).  The only argument for this I'm familiar with involves characteristic classes, but I'm vaguely aware of one using Steenrod squares.
Finally, I just wanted to point out that when one says "all sphere's admit orientation reversing diffeomorphisms", one must be careful to throw out most exotic spheres.  An exotic sphere is a manifold which is homeomorphic to $S^n$ but not diffeomorphic to it (and such things are known to exist, starting in dimension $7$).
It's a fact that for each fixed dimension the set of diffeomorphism classes of smooth manifolds homeomorphic to $S^n$ forms a group under connect sum.  The inverse to an exotic sphere is the same sphere with reveresed orientation.  Hence, exotic spheres admitting orientation reversing diffeomorphisms correspond to order $2$ elements in this group.  Then, for example, in dimension $7$, the group is known to be cyclic of order $28$, so only $1$ of the 13 unoriented types of exotic spheres admits orientation reversing diffeomorphisms (though, being homeomorphic to $S^n$, it admits orientation reversing homeomorphisms!)
A: The $n$-sphere $S^n$ is $\{(x_0, x_1,\ldots ,x_n):\sum {x_i}^2 = 1\}$ The map $f:S^n \to S^n$ sending $(x_0,x_1,\ldots , x_n)$ to $(-x_0,x_1,\ldots , x_n)$ is an oreintation-reversing diffeomorphism.
Sorry I don't have an answer for complex projective spaces.
A: The reason complex projective space $\mathbb{CP}^{2k}$ has no orientation-reversing homeomorphism is because the top dimensional cohomology is generated by an even power of the generator, $x$, of $H^2(\mathbb{CP}^{2k})$. So any self-homeomorphism will send $x$ to $\lambda x$ ($\lambda\neq 0$), and the top cohomology will have $x^{2k}\mapsto \lambda^{2k} x^{2k}$. Since $\lambda^{2k}>0$, this preserves orientation. As Georges pointed out, this argument doesn't work for $\mathbb{CP}^{2k+1}$.
A: Let me add one more argument for $\mathbb{CP}^2$ because it generalizes to many other 4-manifolds and the intuition relies on the intersection form and signature.
Its signature is non-zero, and any orientation reversing map reverses the signature. So if there is an orientation preserving map $f:\mathbb{CP}^2 \to \overline{\mathbb{CP}^2}$, then $\sigma(\mathbb{CP}^2) \mapsto \sigma(\overline{\mathbb{CP}^2})=\sigma(\mathbb{CP}^2)$ and hence $1=-1$.
