Let $F$ be an isometry of the Euclidean space $\mathbb{R}^3$.
Hence $F$ is orthogonoal transform followed by translation by a constant vector.
Let M be a surface of $\mathbb{R}^3$ that is connected, not in any plane.
How to prove that if $F(m)=m$ for all $m \in M$, then $F$ is the identity map?
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You know that $F$ and $I$ are both affine maps. So, to prove they are equal it is enough to prove that they coincide in $n+1=4$ points which are affinely-independent: i.e. such that the smallest affine space containing them is the whole space.
addendum You known that the smallest affine space containing $M$ is the whole space (otherwise it would be contained in an affine $2$-dimensional space), so it must contain at least $4$ affinely independent points.
You know that $F(x)=Ux+b$, where $U$ is a $3\times 3$ orthogonal matrix and $b\in\mathbb R^3$.
What is the set $S$ of fixed points of such an $f$? ($S=\{x:f(x)=x\}$). We have $$ Ux+b=x \quad\Longleftrightarrow\quad (U-I)x=-b. $$
Case I. $U-I$ invertible. Then $S$ has a single element.
Case II. $\mathrm{rank}(U-I)=2$, then $S$ is a straight line or the empty set.
Case III. $\mathrm{rank}(U-I)=1$, then $S$ is a plane or a subset of it.
Case IV. $\mathrm{rank}(U-I)=0$, then $S=\mathrm R^3$, if $b=0$ or $S=\varnothing$ if $b\ne 0$.
Therefore, only Case IV works, and for $b=0$.