The word ``operator'' possibly means a linear mapping from one vector space to itself. Could you please tell us on what space you operator is defined and how it acts.

I can but may be instead of my example you take the one which appeared in the comment of Will Sawin: as the diffeomorphism $R\to R$ you take $x\mapsto x+ x^3$. It has only one fixed or periodic point, this is the point $x=0$; its derivative at $x=0$ is $1$, and it can not be an isometry since isometry preserves volume and therefore can not map the interval [-1, 1] on the interval $[-2,2]$ as the diffeomorphism does.

I agree that the diffeomorphism $x\mapsto x+ x^3$ is possibly the simples counterexample. The observation that the flow of $ x^3 \frac{\partial}{\partial x} $ diverges does not really make any problems since it is well defined on small intervals around $0$

To Q2: in small dimensions (n=2,3) the mapping $g\to R(g)$ is not overdetermined and at least in dimension 2 there always exists a metric with a given $R_{ijkl}$ satisfying (1,2,3)

to @Konrad Voelkel: connections can be given locally by a finite set of functions (say, affine connections by Christoffel symbols) and ``most'' means that the set of connections such that the holonomy is maximal contains an open everywhere dense subset in the say $C^\infty$ topology

It may be that I misunderstood your question. I was thinking about holonomy groups, i.e., the groups generated along all possible paths. The holonomy group is maximal for a generic connection is the sence that one can arbitrary small perturbe the connection such that the holomony group is the maximally allowed in this situation. I gave examples in my answer. But is the case you indeed want that the parallel translation w.r.t. two connection coincides for every loop, my answer is irrelevant. By Ambrose-Singer it may imply that the curvature of two connections coincide at every point.

It depends of course what you understand by compatible. It is parallel, and the corresponding endomorphism is a complex structure, though not the standard one. Alternatively, you make take another symplectic form $-dx_1\wedge dy_1+ dx_2\wedge dx_2$; it corresponds to the standard complex structure

No, since the stereographic projection sends euclidean circles to the circles on the round sphere.