Let $L:S^2\rightarrow S^2$ be a bijective continue map.
Is there exists $L$ such that $\forall x\in S^2 \Rightarrow Lx\neq x$ and $Lx\ne -x$?
I mean that whether $L$ must have fix point?
In fact ,I guess that $S^2$ in space rotating about time should have not fix point , but I can't precisely explain it.
On
It appears that you are wrong. According to Brouwer fixed point theorem, every continuous function from a compact disk to itself has a fixed point.
For your counter-example in mind, the caveat is that it still holds that $L(0)=0$.
On
There cannot exists such function...
PROOF 1 $L(x)\neq x$ implies $L$ is homotopic to antipodal map, and $L(x)\neq -x$ implies $L$ is homotopic to identity map. but $deg$(identity) = $1$, where $deg$(antipodal) = $-1$ for $S^2$.
Proof 2 $L(x)\neq -x$ implies $L$ is homotopic to identity map. But if $L$ has no fixed point then $Lefschetz \ Fixed\ Point\ Theorem$ (https://en.wikipedia.org/wiki/Lefschetz_fixed-point_theorem) implies $\chi(S^2)=0$, but $\chi(S^2)= 2$. SO contradiction.
On
There is no continuous $L:\>S^2\to S^2$ with $L(x)\ne x$ and $L(x)\ne-x$ for all $x\in S^2$. This is easily seen using the hairy ball theorem ("you cannot comb a hedgehog"), that most readers here have heard of. In particular we need neither injectivity nor surjectivity of $L$.
Assume there were such an $L$, and consider an $x\in S^2$. There is a unique shortest great circle arc connecting $x$ with $L(x)\notin\{x,-x\}$. This arc has an outgoing unit tangent vector ${\bf u}=:{\bf u}(x)$ at $x$, and it is easy to prove that $x\mapsto {\bf u}(x)$ is continuous on $S^2$. But according to the hairy ball theorem such a vector field ${\bf u}(\cdot)$ cannot exist on $S^2$.
Consider the map $x\mapsto -x$, if $x=-x$ then $x=0$ which is not in $S^2$. It is true though that there is always a point $x\in S^2$ such that either $Lx = x$ or $Lx=-x$. The proof of that fact is easy once one can prove that the degree of the antipodal map $A:S^2\to S^2$, $A(x)=-x$ is $-1$.
Edit: Let me explain a bit more. The main thing that makes $S^2$ have this property that for any map $L:S^2\to S^2$ there is a point $x$ for which $Lx=x$ or $Lx=-x$ is that the antipodal map $A:S^2\to S^2$, $A(x)=-x$ is of degree $-1$.
Recall the definition of the degree of a map $f:S^n\to S^n$. Let $n\geq 1$ and $f:S^n\to S^n$ a continuous function. We know that $H_n(S^n) = \mathbb{Z} = \langle \alpha\rangle$. Then, $f$ induces a morphism $f_*:H_n(S^n)\to H_n(S^n)$ given by $\alpha\mapsto d\alpha$ for some $d\in\mathbb{Z}$. This $d$ is called the degree of the map $f$ and will be denoted $\deg f$. (Clearly it’s well defined since the only other possible generator is $-\alpha$ and the $f_*(-\alpha) = -f_*(\alpha) = d(-\alpha)$).
Clearly if $f,g:S^n\to S^n$ are homotopic, then $\deg f = \deg g$ (since they induce the same morphism on the homology group!). This means that the degree is a homotopic invariant. It’s also clear that $\deg 1_{S^n}= 1$ (since it induces the identity map on the homology). It’s an easy exercise to verify that $\deg (f\circ g) = \deg f \deg g$.
Notice that if we have $f,g:S^n\to S^n$ such that $f(x)\neq g(x)$ for every $x\in S^n$, we must have that $f \simeq -g$. Indeed, notice that given two points $p,q\in S^n$, the line segment that joins them contains the origin if and only if $p=-q$. Since $f(x)\neq g(x)$ we must have that the origin is not in the line segment between $f(x)$ and $-g(x)$. We thus consider the linear homotopy and normalize it such that it lies on the circle. That is $H:S^n\times I\to S^n$ given by $$H(x,t) = \dfrac{tf(x) + (1-t)(-g(x))}{||tf(x) + (1-t)(-g(x))||}$$ is a homotopy between $f$ and $g$.
We want to prove that every continuous function $L:S^n\to S^n$ with even $n$ must have a point $p\in S^n$ such that either $L(p)=p$ or $L(p)=-p$. Indeed, suppose $L(x)\neq x$ and $L(x)\neq -x$ for every $x\in S^n$. By the previous paragraph we must have that $L\simeq \mathrm{id}$ and $L\simeq -\mathrm{id} = A$. The result will follow on the technical lemma (which can be proved by an astute use of Mayer-Vietoris) that the degree of the antipodal map $\deg A = (-1)^{n+1}$ which will imply that, for even $n$ $$1=\deg\mathrm{id} = \deg L = \deg A = (-1)^{n-1}=-1$$ Thus, there will be a point $p\in S^n$ such that $L(p)=p$ or $L(p)=-p$.
For the case when $n$ is odd, we have $S^{2k-1}\subset\mathbb{R}^{2k}$ and thus a point $p\in S^{2k-1}$ is of the form $p=(x_1,\ldots, x_k, x_{k+1},\ldots,x_{2k})$. Consider the map $L:S^{2k-1}\to S^{2k-1}$ given by $$L(x_1,\ldots ,x_{2k}) = (-x_{k+1},\ldots , -x_{2k}, x_1, \ldots , x_k)$$ and check that it does not have the property (check that neither possibility $Lp=p$ and $Lp = -p$ can happen).