Thanks for all the comments.
Injectivity: Let $p\neq q \in M,$ and let $\gamma:I \to M$ be a geodesic $\gamma(0)=p$ and $\gamma(1)=q.$ Cover $\gamma(I),$ by open sets $U_{\alpha}$ where $f|_{U_{\alpha}}:U_{\alpha} \to f(U_{\alpha})$ is an isometry, by compactness, there is a finite covering of $U_i$'s or equivalently, a partition of $I$ where $0=t_0<t_1<...<t_n=1.$ Thus, $\frac{D}{dt}(\frac{d(f \circ \gamma)}{dt}|_{[t_i,t_{i+1}]})=df (\frac{D}{dt}(\frac{d\gamma}{dt}|_{[t_i,t_{i+1}]}))=df(0)=0,$ therefore, $\frac{D}{dt}(\frac{d(f \circ \gamma)}{dt})$ on $I$ implying that $f \circ \gamma$ is a geodesic in $N$ joining $f(p)$ to $f(q).$ If $f(p)=f(q),$ then $f \circ \gamma$ would be a closed geodesic, contradicting the uniqueness assumption.
Surjectivity: Uniqueness of geodesics joining any two points of $N$ implies that $N$ is complete, then $exp_q:T_qN \to N$ is surjective for any $q \in N.$
Let $p \in M$ be fixed. There is an open neighborhood $U \in M$ containing $p$ s.t. $f|_U : U \to f(U)$ is an isometry. We have $f(exp_p(v))=exp_{f(p)}(df_p(v))$ for all $v \in T_pM.$ Let $q \in N$ be arbitrary. There is a $w \in T_{f(p)}N$ s.t. $exp_{f(p)}(w)=q.$ Since, $df_p$ is an isomorphism of vetor spaces, there is a $u \in T_pM$ s.t. $df_p(u)=w.$ Hence, $ f(exp_p(v))=exp_{f(p)}(df_p(v))=q$ and we’re done.
P.S. Sometimes triviality doesn't show itself, when exhaustion has captured our feelings:)
I'll drop the * compared with your image.
A (unit-speed) geodesic in $M$ is uniquely specified by its starting point and a starting direction vector. We start a geodesic at $p\in N$ and a direction $v\in H$. The isometry fixes $(p,v)$ so fixes the geodesic. So $\exp_p(B(0,r)\cap H)$ is fixed.
The only thing left to prove is that $H\supseteq T_pN$. If $\gamma$ is a smooth curve on $N$ with $\gamma(0)=p$, then $\gamma$ is fixed, so $\gamma'(0)$ is invariant, so $T_pN\subseteq H$.
Best Answer
See Proposition 24 of Chapter 5, section 10 in Peter Petersen's book Riemannian Geometry, which states the following:
Suppose $S \subset \textrm{Iso}(M,g)$ is a set of isometries. Then each connected component of the fixed point set is a totally geodesic submanifold $X \subset M$. Here totally geodesic means that the second fundamental form of $X$ in $M$ is identically zero.
Thus (as is basically indicated in the comments) it makes sense to look at collections of isometries instead of a single isometry.
In this case, if the fixed point set happens to be one dimensional, it will be a geodesic.