Let $A$ be a finite dimensional unitary algebra over the field of real numbers $\mathbb{R}$. Let $\psi : A \to A$ be an arbitrary automorphism. Then the inequality $\psi(t) \neq t + 1$ holds for all $t \in A$.
I have already shown that the statement holds for all inner automorphisms of $A$ by embedding $A$ into the algebra of real $n \times n$ matrices, where the dimension of $A$ is $n$.
I think it will help to clarify at the beginning just what an algebra automorphism such as $\psi$ is. We will accept a definition in accord the widipedia entry Algebra homomorphism; that is, taking $A$ to be an algebra over the arbitrary field $K$:
1.) $\psi$ is a linear operator on $A$, considered a vector space:
$\forall x, y \in A, \; \forall k \in K, \; \psi(ka + b) = k\psi(a) + \psi(b); \tag 1$
2.) $\psi$ preserves the multiplication operation of $A$ in the sense that
$\forall x, y \in A, \psi(xy) = \psi(x) \psi(y). \tag 2$
I denote the unit of $A$ by $1_A$; we note that $\psi$, being an automorphism, is both injective and surjective, whence
$a = \psi(\psi^{-1}(a)) = \psi(1_A \psi^{-1}(a)) = \psi(1_A)\psi(\psi^{-1}(a)) = \psi(1_A)a, \tag 3$
$a = \psi(\psi^{-1}(a)) = \psi(\psi^{-1}(a)1_A ) = \psi(\psi^{-1}(a))\psi(1_A) = a\psi(1_A), \tag 4$
that is
$\psi(1_A)a = a = a \psi(1_A), \; \forall a \in A, \tag 5$
which demonstrates that
$\psi(1_A) = 1_A; \tag 6$
that is, $\psi$ fixes the unit of $A$ $1_A$; it is preserved under the action of any automorphism of $A$.
It further follows from (1) and (2) above that the subalgebra
$K1_A \subset A \tag 7$
is also fixed, element-wise, by $\psi$:
$\forall k \in K, \; \psi(k1_A) = k \psi(1_A) = k1_A, \tag 8$
using (6).
We note in passing that we have as yet made no explicit reference to the real field $\Bbb R$; everything we prove here will hold for $A$ an algebra over an arbitrary field $K$, provided that
$\text{char}(K) = 0; \tag{8.5}$
we will thus continue in this vein.
Suppose now that . $\dim_K A = n < \infty; \tag 9$
then for any
$t \in A, \tag{10}$
the set
$P_t = \{1_A, t, t^2, \ldots, t^n \} \tag{11}$
is linearly dependent over $K$, since
$\vert P_t \vert = n + 1; \tag{12}$
thus there exist
$a_0, a_1, \ldots, a_n \in K, \tag{13}$
not all $0$,
$\exists j, \; 0 \le j \le n, \; a_j \ne 0, \tag{13.5}$
such that
$p(t) = a_n t^n + a_{n - 1}t^{n - 1} + \ldots + a_1 t + a_0 1_A = \displaystyle \sum_0^n a_k t^k = 0, \tag{14}$
where we adopt the customary convention
$t^0 = 1_A. \tag{15}$
We may write (14) in the form
$p(t) = a_n 1_A t^n + a_{n - 1} 1_A t^{n - 1} + \ldots + a_1 1_A t + a_0 1_A = \displaystyle \sum_0^n a_k 1_A t^k = 0, \tag{16}$
from which it is clear we may consider $t$ as satisfying a polynomial
$p(x) = \displaystyle \sum_0^n (a_k 1_A)x^k \in (K1_A)[x], \tag{17}$
and given that there exists such a polynomial, we may affirm that of all the non-zero polynomials in $(K1_A)[x]$ satified by $t$, there exists (at least) one, $m(x)$, of minimal degree:
$m(x) = \displaystyle \sum_0^{\deg m} (m_k 1_A) x^k \in (K1_A)[x], \tag{18}$
$m(t) = \displaystyle \sum_0^{\deg m} (m_k 1_A) t^k = 0. \tag{19}$
Before proceeding further we observe that
$\deg m(x) \ne 0, \tag{20}$
lest
$m_01_A = (m_0 1_A) 1_A = (m_0 1_A)t^0 = 0$ $\Longrightarrow m_0 = 0 \Longrightarrow m(x) = 0, \tag{21}$
forbidden by our hypotheses on $m(x)$.
Now suppose that
$\exists t \in A, \; \psi(t) = t + 1_A; \tag{22}$
we note in passing that this assumption rules out the possibiity that
$\deg m(x) = 1, \tag{23}$
lest again that
$m_1 1_A t + m_01_A = 0 \Longrightarrow m_1 1_A t = -m_0 1_A$ $\Longrightarrow t = -m_1^{-1}m_0 1_A \in K1_A; \tag{24}$
but in accord with (6), $\psi$ fixes $1_AK$, so (22) becomes impossible in this case and thus
$\deg m(x) \ge 2. \tag{25}$
We now apply the automorphism $\psi$ to (19) and find
$0 = \psi(m(t)) = \psi \left (\displaystyle \sum_0^{\deg m} (m_k 1_A) t^k \right ) = \displaystyle \sum_0^{\deg m} \psi((m_k 1_A)t^k)$ $= \displaystyle \sum_0^{\deg m} \psi(m_k 1_A)\psi(t^k) = \displaystyle \sum_0^{\deg m} (m_k1_A) (\psi(t))^k= \displaystyle \sum_0^{\deg m} (m_k1_A) (t + 1_A)^k; \tag{26}$
we expand $(t + 1)^k$ via the ordinary binomial theorem:
$(t + 1)^k = t^k + kt^{k - 1} + \text{terms of lower degree}, \tag{27}$
so if we gather the terms of degree $\deg m$ and $\deg m - 1$ from (26) and separate them out we obtain
$\displaystyle \sum_0^{\deg m} (m_k1_A) (t + 1_A)^k$ $= m_{\deg m}1_A t^{\deg m} + (m_{\deg m - 1} + \deg m)1_At^{\deg m - 1} + \displaystyle \sum_0^{\deg m - 2}(m_k 1_A)(t + 1_A)^k = 0; \tag{28}$
next, writing (19) in the form
$m(t) = m_{\deg m}1_A t^{\deg m} + m_{\deg m - 1}1_A t^{\deg m - 1} + \displaystyle \sum_0^{\deg m - 2} (m_k 1_A) t^k = 0, \tag{29}$
we discover upon subtracting these two equations that
$\deg m 1_A t^{\deg m - 1} + \text{(terms of lower degree in} \; t) = 0; \tag{30}$
however, (30) affirms that $t$ satisfies a polynomial of degree less than that of $m(x)$, $\deg m$; from this contradiction we infer that (22) cannot bind, i.e. that
$\forall t \in A, \; \psi(t) \ne t + 1_A. \tag{31}$
Nonexistence of $\psi(t) = t + 1$ finite dimensional algebra automorphisms