Let $n\geq 2$ and a bijection $T:\ \mathbb R^n\longrightarrow\mathbb R^n$ satisfies $\ T(0)=0\,$ and $\,T$ maps straight lines to straight lines.
Then $T$ is a linear map.
My proof :
Let $(P)\subset\mathbb R^n$ be a 2-plane and the straight lines $d_1,d_2,d_3\subset(P)$ which intersect pairwise respectively at $A,B,C$. Since $T$ is bijective and $T(d_1),T(d_2),T(d_3)$ are straight lines, they intersect pairwise respectively at $T(A)=A',\,T(B)=B',\,T(C)=C'$. Hence they form a 2-plane $(A'B'C')$.
For each straight line $d\subset (P)$, if $d$ doesn't contain $A,B,C$ then $d$ must intersect with at least 2 lines of $d_1,d_2,d_3$ at at least 2 points. Since $T$ is bijective, so is $T(d)$. Hence, $T(d)\subset (A'B'C')$. We conclude $T(P)\subseteq(A'B'C')$. By a similar deduce, we show that $T^{-1}(A'B'C')\subseteq(P)$, therefore $T(P)=(A'B'C')$.
Let $d,d'\subset(P)$ be 2 distinguish parallel lines. Since $T(d)\cap T(d')=\varnothing$ and $T(d),T(d')\subset T(P)$, we have $T(d), T(d')$ are parallel. Thus, if $A,B,C,D$ are the vertices of the parallelogram $ABCD$, then $A',B',C',D'$ must be the vertices of the parallelogram $A'B'C'D'$.
Let $u,v\in\mathbb R^n\setminus{0}$ be linearly independent. Since $0,u,u+v,v$ be the vertices of a parallelogram, so do $0,T(u),T(u+v),T(v)$, this implies $T(u+v)=T(u)+T(v)$.
Let $\mu,\lambda\in\mathbb R\setminus\{0\}$ and $u,v\in \mathbb R^n\setminus\{0\}$ be linearly independent. Then $\{\mu u,v\},\, \{\lambda u,\mu u+v\},\, \{\lambda u+\mu u,v\}$ are also linearly independent. Thus, \begin{align} T(\lambda u)+T(\mu u)+T(v)\,&=\,T(\lambda u)+T\big(\mu u+v\big) \\ &=\,T\big(\lambda u+\mu u+v\big) \\ &=\,T\big(\lambda u+\mu u\big)+T(v),\tag1 \end{align} which implies $T\big(\lambda u+\mu u\big)=T(\lambda u)+T(\mu u)$. So, $T$ is additive.
Given $0\ne u\in\mathbb R^n$. Because $\langle u\rangle$ is a straight line, $T(\langle u\rangle)$ is also straight line. In the other hand, $\big<T(u) \big>$ is also a straight line, and both $T(\langle u\rangle),\,\big<T(u) \big>$ contain $0, T(u)$, therefore $T(\langle u\rangle)=\big<T(u) \big>$. We deduce \begin{align} \forall\lambda\in\mathbb R,\ \exists\,\mu(u,\lambda)\in\mathbb R:\ T(\lambda u)=\mu(u,\lambda) T(u).\tag2 \end{align} Given $\lambda,\varepsilon\in\mathbb R\setminus\{0\}$ and $u,v\in \mathbb R^n\setminus\{0\}$ be linearly independent. We have the straight lines $\big[u,v\big]$ and $\big[\lambda u,\lambda v \big]$ are parallel, and the straight lines $\big[\varepsilon u,v\big]$ and $\big[\lambda\varepsilon u,\lambda v \big]$ are parallel. Hence, $\big[T(u),T(v)\big]$ and $\big[T(\lambda u),T(\lambda v)\big]$ are parallel, and $\big[T(\varepsilon u),T(v)\big]$ and $\big[T(\lambda\varepsilon u),T(\lambda v)\big]$ are also parallel. By Thales's Theorem, we have \begin{align} \frac{\big\|T(\lambda u)\big \|}{\big\|T(u) \big\|}=\frac{\big\|T(\lambda v)\big \|}{\big\|T(v) \big\|}=\frac{\big\|T(\lambda\varepsilon u)\big \|}{\big\|T(\varepsilon u) \big\|},\tag3 \end{align} thus \begin{align} \big|\mu(u,\lambda)\big|\,=\,\big|\mu(\varepsilon u,\lambda) \big|\tag4 \end{align} or \begin{align} \mu(u,\lambda)\,=\,\pm\mu(\varepsilon u,\lambda).\tag5 \end{align}
Now, we have \begin{align} \mu(u,\lambda\varepsilon)T(u)\,=\,T(\lambda\varepsilon u)\,=\,\mu(\varepsilon u,\lambda)\mu(u,\varepsilon)T(u)\tag6 \end{align} thus \begin{align} \mu(u,\lambda\varepsilon)\,=\,\mu(\varepsilon u,\lambda)\mu(u,\varepsilon)\,=\,\pm\mu(u,\lambda)\mu(u,\varepsilon)\tag7 \end{align}
We also have \begin{align} \mu\big(u,\lambda+\varepsilon\big)T(u)\,&=\,T\big((\lambda+\varepsilon)u \big) \\ &=\,T(\lambda u)+T(\varepsilon u) \\ &=\, \big[\mu(u,\lambda)+\mu(u,\varepsilon)\big] T(u)\tag8 \end{align} hence $\mu\big(u,\lambda+\varepsilon\big)=\mu(u,\lambda)+\mu(u,\varepsilon)$.
We are so close to show that $\mu(u,\cdot)$ is a field automorphism of $\mathbb R$. And a well known result states that the only field automorphism of $\mathbb R$ is the identity map, therefore $\mu(u,\cdot)=id$. This will complete our lengthy proof. But, my final impedent, how to show that $\mu(u,\lambda\varepsilon)=\mu(u,\lambda)\mu(u,\varepsilon)$, which means to show that the negative case is no true ? Since I have tried and can't find out any contradiction for this case.
May anyone give me some helps ? If you get another fascinating proof for this Theorem, like using calculus and try showing $T$ is continuous some way for instance, feel free to share your great idea. Thanks.
The part of the proof where you use the Thales theorem should also give you the sign.
If $[T(u), T(v)]$ and $[T(\lambda u), T(\lambda v)]$ are parallel, then not only $\frac{\| T(\lambda u) \|}{\| T(u) \|} = \frac{\|T(\lambda v)\|}{\|T(v)\|}$, but also $T(\lambda u)$ and $T(\lambda v)$ lie at the same side of the line $[T(u), T(v)]$. In terms of $\mu$ this means not only $|\mu(u,\lambda)| = |\mu(v, \lambda)|$, but $\mu(u,\lambda) = \mu(v, \lambda)$.