Let me add my partial answer. Let us take $B$ to be upper triangular invertible $3\times 3$ matrices over a finite field of $p$ elements, $p\geq 5$ an odd prime. Then I claim that the automorphisms group is generated by the inner ones and the map $\sigma: g\mapsto \kappa ^tg \kappa ^{-1}$ where $\kappa$ is the longest Weyl group element in $GL_3$. Let $\theta$ be an automorphism. It takes $U$ to $U$, where $U$ is the unipotent radical of $B$ (since $U$ is the unique $p$-sylow subgroup).
After changing $\theta$ an inner conjugation if necessary, we may assume that $T$ goes to $T$ where $T$ is the group of diagonals: The diagonals contain an $l$-Sylow subgroup of $B$ where $l$ is a prime dividing $p-1$ and we may assume $\theta $ stabilises this $l$-Sylow subgroup $H$. Therefore, $\theta $ stabilises the centraliser of this $H$. But the centraliser is $T$.
Since $p\geq 5$ we may take logarithms, and the automorphism $\theta $ takes one root space into another. The root group $X_{13}$ being central, goes into itself. The other two $simple$ root groups $X_{12}$ and $X_{23}$ may be permuted. After changing $\theta $ by $\sigma$ if needed, one can see that $\theta $ stabilises each root group, and after changing $\theta $ by an inner conjugation by an element of $T$, we see that each simple root group is left point-wise fixed by $\theta$.
[Remark]: I think this argument extends to $n\times n$ matrices as well , at least when $p>n$ (we took logs). Perhaps with some more group theory tricks, we can avoid logarithms and get this for all $n$ and all odd primes $p$.