Is it true that if $f(p)=p$ and d$f_p=\operatorname{id}$ then $f(\exp_p(v))=\exp_p(v)\ \forall v\in T_pM$?

Question

$M$ is a Riemannian manifold, $f$ a (smooth) function from $M$ to itself, $v$ only vectors for which the exponential map is defined.

Then, Is it true that

if $f(p)=p$ and $\mathrm{d}f_p=\operatorname{id}$ then $f(\exp_p(v))=\exp_p(v)\ \forall v\in T_pM$?

Answer 1

The answer is no. For example take the function: $f(x)=x^2+x$. $f(0)=0,f'(x)=2x+1 \Rightarrow f'(0)=1$, however $\text{exp}_0(v)=v$, so $f(\text{exp}_0(v))=f(v)=v^2+v \neq v = \text{exp}_0(v)$.

Note:

Any Riemannian isometry whose domain is a connected manifold, is determined by its value and its differential at a single point.

So, if you assume $f$ is an isometry, then it follows that $f=Id$, so the claim is true.

This is proved for instance in Do-carmo's book "Riemannian Geometry". (see pg 163 lemma 4.2).