Is there a compact Riemanian manifold $M$ not diffeomorphic to sphere or real or complex or quaternion projective space which admit a diffeomorphism $f$ with the property that $$\forall x \in M, \quad d(f(x), x)=diam(M)$$ where $d$ is the metric arising from the Riemannian metric and diameter of $M$ with repect to this metric is denoted by $diam(M)$

4$\begingroup$ It's probably not what you were looking for, but the flat torus (which is $S^1 \times S^1$) will also have this property as well. In general, products of spheres and projective spaces will have diffeomorphisms satisfying this equation. $\endgroup$– Gabe KSep 11 '18 at 22:16

$\begingroup$ @GabeK Yes that is right thank you. But what is an example of a compact manifold which does not satisfy the fixed point property does not admit such diffeomorphism? $\endgroup$– Ali TaghaviSep 11 '18 at 22:25
The article by X. Liu and Sh. Deng "The antipodal sets of compact symmetric spaces" gives many examples, e.g. $\mathrm{SU}(2n)$, $\mathrm{Spin}(5)$, $\mathrm{Spin}(7)$,.... All those spaces have unique antipodal points which implies that the antipodal map is a diffeomorphism, see below. Here the antipodal points at $p$ are defined as $A(p)=\{ x\in M ~~d(x,p)=\mathrm{diam}M\}$.
If a homogeneous space, equipped with a leftinvariant metric has unique antipodal points then by compactness one gets a homeomorphism $f$ with the desired properties. By leftinvariant of the metric the map $t\mapsto d(\exp(tX)x_0,\exp(tX)y_0)$ is constant so that $f(\exp(tX)x) = \exp(tX)f(x)$ which shows $f$ is differentiable. Here $\exp(X)$ denotes the isometry generated by the Killing field $X$.

$\begingroup$ Thank you very much for your very interesting and perfect answer. $\endgroup$ Sep 12 '18 at 21:18