In my book ist says:
(Fundamental Theorem of Affine Transformations): Given two ordered sets of three non-collinear points each, there exists a unique affine transformation f mapping one set onto the other.
The proof for the existence is given, but I am not quite sure how to prove the uniqueness.
I made the following attempt:
First we define the affine transformation f, that maps the unit vectors {0,i,j} to a triple of non collinear points (p,q,r}. Let h and h' be two affine transformations that map the triple of non collinear points {p,q,r} to another triple of non collinear points {p',q',r'}. We can consider the composition h∘f and h'∘f (that are also affine transformations, since the composition of two affine transformations is also affine), that map {0,i,j} to {p',q',r'}.
From this point on I am not sure how to proceed. I read somewhere else that we can prove that h and h' are the same transformation, since an affine transformation is uniquely determined by its effect on the points. But this is nowhere mentioned in the book I am using.
If the ambient space is an affine plane, then there is uniqueness. Otherwise, in dimension $>2$, the affinity is not unique.
Let $\mathcal A(E)\;$ be an affine space over the vector space $E$ of dimension $n$, and
$$ P_0,\;P_1,\ldots,\;P_n \qquad \text{and} \qquad Q_0,\;Q_1,\ldots,\;Q_n $$
two $n+1$-tuples each consisting of affine independent points of $\;\mathcal A(E)$. That means that
$$ P_1-P_0,\;P_2-P_0,\ldots,\;P_n-P_0 \qquad\text{and}\qquad Q_1-Q_0,\;Q_2-Q_0,\ldots,\;Q_n-Q_0 $$
are two $n$-tuples of linearly independent vectors of $E$, i.e. two bases of $\;E$.
It is well known that there is a bijective linear function $\;L\;:\;E\to E\;$ which transform the first basis into the second, that is
$$ L(P_i-P_0) = Q_i-Q_0 \qquad\quad (i=1,\ldots,n). $$
The function
$$ f\;:\;{\mathcal A}(E)\to {\mathcal A}(E)\qquad\qquad f(P):= Q_0+L(P-P_0) \tag{*}$$
is therefore an affinity (because $L$ is bijective) such that
$$f(P_i) = Q_0+L(P_i-P_0)= Q_0+(Q_i-Q_0)=Q_i\qquad\quad (i=1,\ldots,n), $$
which obviously also applies for $\;i=0$.
The linear function $\;L\;$ in $\;(^*)$, i.e. the linear part of $f$, is often denoted by $\;df$, so we can write $\;f(P)=Q_0+df(P-P_0)$, where $Q_0=f(P_0)$.
$\raise5ex{}$ Uniqueness$\quad$ Suppose that there is another affinity $\;g\;:\;\mathcal A(E)\to\mathcal A(E)\;$ such that
$$ g(P_i)=Q_i\qquad\quad(i=0,1,\ldots,n). $$
We have:
\begin{align} d(g\circ f^{-1})(Q_i-Q_0) &= dg\circ d(f^{-1})(Q_i-Q_0) =\\[1.5ex] &= dg(P_i-P_0)=\\[1.5ex] &= g(P_i)-g(P_0) =\\[1ex] &= Q_i-Q_0 \end{align}
and therefore $\;d(g\circ f^{-1})=1_E$, since the family $\;(Q_i-Q_0)_{i=1,\ldots,n}\;$ is a basis of $\;E$. It follows
$$ g\circ f^{-1}(Q)=g\circ f^{-1}(Q_0)+1_E(Q-Q_0)=Q_0+(Q-Q_0)=Q $$
i.e. $\;g\circ f^{-1}=1_{\mathcal A(E)}$, and finally $\;g=f$.