unique minimal models and birational automorphisms of algbraic surfaces In his book on cubic forms Manin writes that if an algebraic surface has unique minimal
model $X$ then "it follows easily from definitions" that the group of
automorphisms of the field $k(X)$ is isomorphic to the group of
biregular automorphisms of $X$. 
it is ch. III, 18.7.1. (Here is a link to Google books)
While an automorphism of the field surely induces a birational map
(an isomorphism of dense open subsets) $f$ from $X$ to itself, I fail to see how one can produce a birational
morphism (a totally defined morphism that induces a birational map after restriction to a dense open subset)  from $X$ to itself out of it. By closing the graph of $f$
in $X \times X$ one can find a modification of $X$, $\varphi: X'\to X$,
and factor $f$ as a composition of a birational map $X' \to X$ (projection of the closure of the graph of $f$ on the second factor $X$) and the inverse
of a the modification $\varphi$. It is unclear though how this
factorisation helps, and how one can use uniqueness of minimal models
to relate $f$ to a biregular morphism of $X$.
Related MO question and answers: https://mathoverflow.net/questions/120372/when-does-autx-birx-hold/
(note in particular the answer of Christopher Liedke, though it is about birational morphisms; it is also unclear to me how this answer is related to uniqueness of minimal models mentioned by Francesco Polizzi in his  comment)
 A: Let $X$ be the unique minimal surface in its birational class. Let $f$ be a birational self map (which, as you said, corresponds to some automorphism of $k(X)$). We want to show it is a biregular morphism. Since $X$ is smooth, $f$ is defined off a codimension 2 closed subset. Therefore, it is not defined at isolated points. One can resolve the map blowing them up (if you are not familiar with this, Beauville's book is a very good reference). Thus, we have a birational surface $Y$ with morphisms $p: Y \rightarrow X$ and $q: Y \rightarrow X$ such that $q=f \circ p$. Assume that $Y$ is the minimal resolution of the morphism $f$. In particular, all the $p$-exceptional curves are not $q$-exceptional.
If $f$ is not a morphism, we need to blow up some points, and $f$ is not the identity. Therefore, there are some $p$-exceptional curves. Also, the $p$-exceptional curves are curves with negative self intersections (the last one appearing being $(-1)$-curves). 
Now, the fact that $X$ is the unique minimal surface in its birational class implies that $K_X$ is nef (see Theorem 1.29 in Kollàr-Mori). Let $E$ be a $p$-exceptional curve that is also a $(-1)$-curve. Then, $K_Y \cdot E = -1 < 0$. Then, the image of $E$ under $q$, call it $C$, is a curve such that $K_X \cdot C < 0$. This happens because at worst we are contracting some curve that meets $E$. More precisely, we have
$$
K_Y= q^* K_X + \sum a_i F_i,
$$
where $F_i \neq E$ for all $i$. Furthermore, since $X$ is smooth, $a_i > 0$ for all $i$. Now, we want to compute $K_X \cdot q_* E$ and show it is negative. Then, we have
$$
K_X \cdot q_* E = q^* K_X \cdot q^*q_* E = q^* K_X \cdot E = (K_Y - \sum a_i F_i) \cdot E \leq-1.
$$
The first equality always holds on surfaces; the second one is true since $q^*q_* E = E + \sum b_i F_i$, but the $F_i$'s are $q$-exceptional and $q^*K_X$ does not see $q$-exceptional stuff; the inequality follows from the fact that $K_Y \cdot E =-1$, the coefficients $a_i$ being positive and $E \cdot F_i \geq 0$ (since $E \neq F_i$).
Since $K_X$ is nef, this can not happen, and so $f$ has to be an isomorphism, i.e. $f$ is a morphism.
Arguing similarly with the inverse of $f$ and their composition, you can see $f$ is actually biregular.
If $X$ is not the unique minimal surface in its birational class, then $K_X$ is not nef, and we would not get the above contradiction. For instance, consider the minimal surfaces in the birational class of $\mathbb{P}^2$: These are $\mathbb{P}^2$ itself, and the Hirzebruch surfaces $\mathbb{F}_n$ for $n \neq 1$ (for $n=1$ it is the blow-up of $\mathbb{P}^2$ at one point).
Consider $\mathbb{F}_0 = \mathbb{P}^1 \times \mathbb{P}^1$. Blow up a point $p \in \mathbb{F}_0$. For each ruling (horizontal and vertical respectively) there is a unique element through $p$: call them $F_1$ and $F_2$ respectively. Now, on the blow-up of $\mathbb{F}_0$ at $p$ there are three $(-1)$-curves: the exceptional curve $E$, and the strict transforms of $F_1$ and $F_2$. Then, you can blow down both strict transforms, and you end up with $\mathbb{P}^2$. Similarly, you can start with $\mathbb{P}^2$, blow up two points on a line, then contract the strict transform of the line, and end up with a copy of $\mathbb{F}_0$. If you do these two operations in order, and make your choices generic (i.e. the second step is not undoing the first one, just choose another line), you get a birational self map on $\mathbb{F}_0$. On the other hand, this map contracts two lines and extracts two other lines. So, it can not be extended to a morphism.
The uniqueness of minimal model exactly guarantees that the above pathology does not happen: Morally, you can not replace a rational curve with another.
