This question is related to this one. In that question, it is stated that nilpotent elements of a non-commutative ring with no non-trivial ring automorphisms form an ideal. Ted asks in the comment for examples of such rings but there are no answers. I would also like to know whether there are such rings and hence this question.
[Math] non-commutative ring with a trivial automorphism group
noncommutative-algebraring-theory
Best Answer
I think I have one. Let $k$ be the field with $2$ elements. Let $R$ be the $k$-algebra with generators $x$, $y$ and $z$, modulo the relations $$zx=xz,\ zy=yz,\ yx=xyz.$$ It is not hard to see that monomials of the form $x^i y^j z^k$ are a basis for $R$. We will call these the standard monomials.
For any $f \neq 0$ in $R$, write $f = \sum f_{ij}(z) x^i y^j$. We will define the leading term of $f$ to be the term $f_{ij}(z) x^i y^j$ where we choose $i+j$ as large as possible, breaking ties by choosing the largest possible power of $i$.
Lemma The center of $R$ is $k[z]$.
Proof Let $Z$ be central and write $Z$ in the basis of standard monomials. Since $Zx=xZ$, there are no powers of $y$ in $Z$. Since $Zy=yZ$, there are no powers of $x$ in $Z$. $\square$.
Lemma Every automorphism of $R$ acts trivially on the center of $R$.
Proof Every automorphism of $k[z]$ is of the form $z \mapsto az+b$ for $a \in k^{\ast}$. Since $k$ has two elements, we must have $\sigma: z \mapsto z+b$. Any automorphism of $R$ descends to an automorphism of the abelianization, which is $k[x,y,z]/(xy(z-1))$. Since $z-1$ is a zero divisor in the abelianization, $z+b-1$ must be a zero divisor as well, and this forces $b$ to be zero. $\square$.
Lemma If $f$ and $g$ have leading terms $f_{ij}(z) x^i y^j$ and $g_{kl}(z) x^k y^l$, then the leading term of $fg$ is $z^{jk} f_{ij}(z) g_{kl}(z) x^{i+k} y^{j+l}$.
Proof A computation. $\square$
Now, suppose that we have an automorphism $x \mapsto X$, $y \mapsto Y$, $z \mapsto z$ of $R$. Let the leading terms of $X$ and $Y$ be $f(z) x^i y^j$ and $g(z) x^k y^l$.
Lemma The vectors $(i,j)$ and $(k,l)$ are linearly independent.
Proof We are supposed to have $YX=zXY$. Taking leading terms $$z^{il} f(z) g(z) x^{i+k} y^{j+l} = z^{jk+1} f(z) g(z) x^{i+k} y^{j+l}.$$ So $il-jk=1$ and $\det \left( \begin{smallmatrix} i & j \\ k & l \end{smallmatrix} \right)=1$. $\square$
Consider the images $z^a X^b Y^c$ of the standard monomials. Their leading terms are $z^{a+b+c} f(z)^b g(z)^c x^{bi+ck} y^{bi+cl}.$ Using the above lemma, these leading terms are all distinct. So there is no cancellation of leading terms in any sum $\sum s_{abc} z^a X^b Y^c$. So we see that every element in the image of the automorphism must have a leading term of the form $h(z) x^{bi+ck} y^{bj+cl}.$
But automorphism are surjective! So $(1,0)$ and $(0,1)$ must be positive integer combinations of $(i,j)$ and $(k,l)$. So either $(i,j) = (1,0)$ and $(k,l) = (0,1)$ or viceversa. We see that $X$ and $Y$ are of degree $1$ in $x$ and $y$. From this point, it's an easy computation.