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 $x \mapsto ax$) .
 A: 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.
A: Hint: Note that for $a,x \in A$, we have
$$
L_{\varphi(a)}(x) = \varphi(a)x = \varphi(a) \varphi (\varphi^{-1}(x)) = 
\varphi(a\varphi^{-1}(x)) = [\varphi \circ L_a \circ \varphi^{-1}](x)
$$
So, the linear maps $L_{\varphi(a)}$ and $L_a$ must be similar.  Note that $L_{a + 1} = L_a + \operatorname{id}_A$.

There are two approaches that I had in mind with this hint.
The intuitive approach is to consider the eigenvalues of these transformations.  In particular, we note that $\lambda$ is an eigenvalue of $L_a$ if and only if $\lambda + 1$ is an eigenvalue of $L_{a + 1} = L_a + \operatorname{id}_A$.  Because our field has characteristic zero and the set of eigenvalues is finite, it is impossible for $\{\lambda: \lambda  \in \sigma(L_a)\}$ and $\{(\lambda + 1):\lambda \in \sigma(L_a)\}$ to be the same set.  Thus, the operators have distinct spectra and cannot be similar.
The more elegant approach, I think, is to use the trace.  In particular, we note that there exists a "trace operator" over the set of linear transformations on any finite dimensional vector space (there are various equivalent definitions of the trace; any definition will do here).  We note that if $L_a$ were similar to $L_{a+1}$, then we must have $\operatorname{Tr}(L_a) = \operatorname{Tr}(L_{a+1})$.  However, we have
$$
\operatorname{Tr}(L_{a+1}) = \operatorname{Tr}(L_a + \operatorname{id}_A) = \operatorname{Tr}(L_a) + \operatorname{Tr}(\operatorname{id}_A) = \operatorname{Tr}(L_a) + \dim(A)
$$
Because our field has non-zero characteristic and $A$ is non-zero, we must have
$$
\operatorname{Tr}(L_{a+1}) = \operatorname{Tr}(L_a) + \dim(A) \neq \operatorname{Tr}(L_a)
$$
Thus, the two operators cannot be similar.
A: 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 $\varphi(c_1+2c_2+...(n-1)c_{n-1}-1)=c_1+2c_2+...(n-1)c_{n-1}-1$.
