Let $R$ be a ring with identity such that the identity map is the only ring automorphism of $R$. Prove that the set $N$ of all nilpotent elements of $R$ is an ideal of $R$.
I'm asking the question contained in this post $(1)$ in search of the answer. I looked at the hints in $(1)$ and am attempting to carry them forward. Also, please don't refer me to this post ($2)$ because that user embeds $R$ into $R \times \mathbb{Z}$ which is not how I plan to approach this problem. Lastly, note that this post $(3)$ supposes $R$ is commmutative, which I am not.
To prove this, note that every unit $u$ in $R$ is central. To see this, suppose not. Let $u \in R$ be a unit that is non-central. Then there exists $w \in R$ such that $uw \neq wu$. Then $u \neq wuw^{-1}$. Hence, there is an inner automorphism $\varphi$ of $R$ such that $\varphi$ does not fix $u$. Then $\varphi$ is not the identity map, a contradiction. So $u$ is central.
Next, we know that if $u$ is a unit and $v$ is nilpotent, then $u \pm v$ is a unit. It follows that $1-v$ is a unit for every $v \in N$. Now, we must show that $N$ is an ideal, i.e. that $rv$ and $vr$ are in $N$ for all $r \in R$. As $1-v$ is a unit, $(1-v)^{-1} = j$ exists. So
$$(1-v)(1-v)^{-1} = (1-v)j=j-vj=1$$
$$\iff j-1 = vj$$
Now, I'm not sure where to go.
