How can we show that $\varphi(a) \neq a +1$ for every $a \in A$? Let $A$ be a nonzero finite dimensional unital algebra over a field $F$ with
$char(F) = 0$, and let $\varphi$  be an automorphism of $A$. 


  How can we show  that $\varphi(a)  \neq a +1$  for
  every $a \in A$?


How can we prove by $L_{\varphi (a)}$ through $\varphi$,$\varphi^{-1}$, and $L_{a}$?

( $L_{a}$  is  left multiplication map $x \mapsto ax$) .
Now it's on MathSE, let me post the answer as I thought of it.

Suppose $\varphi(a) = a + 1$. Since $A$ is finite-dimensional, $a$ has a minimal (monic) polynomial $P$. Since $\varphi(a)=a+1$, $\varphi$ preserves $F[a]$. Since it maps $0=P(a)$ to $P(a+1)$, we have $P(a+1)=0$, i.e. $Q(a)=0$, where $Q(t)=P(t+1)$. By uniqueness of the minimal monic polynomial, we have $Q=P$. So $P(t)=P(t+1)$. 

It's standard to check that this implies, in characteristic zero, that $P$ is constant ($\star$), hence $P=1$. Clearly 1 cannot be the minimal polynomial of any element, and we have a contradiction.

($\star$) if $P(t)=P(t+1)$ in the polynomial ring $F[t]$, $n$ is a root of $P-P(0)$ for every integer $n\ge 0$, and since in characteristic zero these form infinitely many roots, we deduce $P-P(0)=0$, i.e., $P$ is constant.
Since $A$ is finite dimensional, there exists $n$ such that $a+n=c_0a+c_1(a+1)+c_{n-1}(a+n-1)$, we deduce that $1=c_0a+..+c_{n-2}(a+n-2)+(c_{n-1}-1)(a+n-1)$ where $c_0,...,c_{n-1}\in F$. 

$\varphi(1)=1=\varphi(c_0a+..+c_{n-2}(a+n-2)+(c_{n-1}-1)(a+n-1))=c_0(a+1)+...+c_{n-2}(a+n-1)+(c_{n-1}-1)(a+n)=c_0a+..+c_{n-2}(a+n-2)+(c_{n-1}-1)(a+n-1)+c_0+c_1+..+c_{n-1}=1+c_0+...+c_{n-1}.$

We deduce that $c_0+c_1+...+c_{n-1}=0$, since  $1=c_0a+..+c_{n-2}(a+n-2)+(c_{n-1}-1)(a+n-1)$ we deduce that $(c_0+...+c_{n-1})a+c_1+2c_2+...(n-1)c_{n-1}-a=1$, this implies that $a=c_1+2c_2+...(n-1)c_{n-1}-1$ contradiction, since $\varphi(c_1+2c_2+...(n-1)c_{n-1}-1)=c_1+2c_2+...(n-1)c_{n-1}-1$.