If the definition of Szaro corresponds to the definition of Wikipedia, then the affine structure of $M$ is defined by a connection $\nabla$ whose curvature and torsion forms vanish identically. An affine transformation of $(M,\nabla)$ is a diffeomorphism which preserves the connection $\nabla$. When $M$ is complete, it is the quotient of $\mathbb{R}^n$ by a subgroup $H$ of affine transformation which acts properly and freely on $R^n$. In this case the group of affine transformations of $(M,\nabla)$ is the quotient of the normalizer of $H$ in $\operatorname{Aff}(\mathbb{R}^n)$ by $H$. When $(M,\nabla)$ is compact and complete, I have shown in my thesis that the connected component of the group of affine transformations of $(M,\nabla)$ is nilpotent. This implies that the connected component of $G$ is nilpotent in this case.

To show that the connected component of $\operatorname{Aff}(M,\nabla)$ (the group of affine transformations of $(M,\nabla)$) is nilpotent when $(M,\nabla)$ is compact and complete, one remarks firstly that if $M$ is the quotient of $\mathbb{R}^n$ by the group of affine transformations $H$, then the action of $H$ is irreducible, that is, does not preserve a proper affine subset. To see this, suppose that $H$ preserves an affine subset $V$, then $V/H$ and $M$ are $K(\pi,1)$-Eilenberg McLane space. Thus the cohomology of $M$ and $V/H$ is the cohomology of $H$, since $M$ is compact. (We can assume $M$ is oriented up to a 2-cover.) Thus $H^n(M,\mathbb{R})=H^n(\pi,\mathbb{R})\neq 0$. This implies that $H^n(V/H,\mathbb{R})\neq 0$, thus the dimension of $V/H$ is at least $n$.

Next, you remark that $N(H)_0$ the connected component of the normalizer of $H$ in $\operatorname{Aff}(\mathbb{R}^n)$ commutes with $H$. You deduce that this implies that it acts freely on $\mathbb{R}^n$ since the action of $H$ on $\mathbb{R}^n$ is irreducible.
Now, let $\operatorname{aff}(M,\nabla)$ be the Lie algebra of $\operatorname{Aff}(M,\nabla)$. For every vector $X,Y\in \operatorname{aff}(M,\nabla)$, $\nabla_XY$ is again in $\operatorname{aff}(M,\nabla)$. Since $\operatorname{Aff}(M,\nabla)$ is the quotient of $N(H)$ by the discrete group $H$, their Lie algebras are isomorphic. Remark that the product induced by $\nabla$ on $n(H)$ the Lie algebra of $N(H)$ is the associative product of $\operatorname{aff}(\mathbb{R}^n)$ the Lie algebra of $\operatorname{Aff}(R^n)$ defined by $(A,a).(B,b)=(AB,A(b))$. The associative algebra $n(H)$ does not have an idempotent $(C,c)$ because if $(C,c)$ is such an idempotent, $(C,c).(C,c)=(C^2,C(c))=(C,c)$. This implies that the 1-parameter group generated by $(C,c)$ fixes $-c$. This is in contradiction with the fact that the action of $N(H)_0$ on $\mathbb{R}^n$ is free. An associative algebra which doesn't have a nilpotent element is nilpotent so $n(H)$ is nilpotent, hence $\operatorname{aff}(M,\nabla)$ and $\operatorname{Aff}(M,\nabla)_0$ are nilpotent.

When $(M,\nabla)$ is compact and complete, I have also shown that the natural action of $\operatorname{Aff}(M,\nabla)_0$ on $M$ is locally free, thus its orbits define on $M$ a foliation.

The typical example is the torus $T^n$ which is the quotient of $\mathbb{R}^n$ by the group generated by $n$-translations $t_{e_1},...,t_{e_n}$ such that $(e_1,...,e_n)$ is a basis of $\mathbb{R}^n$. The connected component of $\operatorname{Aff}(T^n)$ is $T^n$. Remark that $\operatorname{Aff}(T^n)$ is not nilpotent although its connected component is nilpotent since the action of $\operatorname{GL}(n,\mathbb{Z})$ on $\mathbb{R}^n$ induces an action of $\operatorname{GL}(n,\mathbb{Z})$ on $T^n$.

Tsemo Aristide

*Dynamique des variétés affines.* **Journal of the London Mathematical Society** 63.2 (2001): 469-486.