Characteristic subgroups of the group of unipotent matrices

Question

Let $\mathbf{U}(n,p)$ be the group of $n\times n$ upper triangular matrices over $\mathbf{F}_p$, with $1$ all over the diagonal.

Let me number the paris $(i,j)$ , $1\leq i<j\leq n$ as follows (note that I am just following the upper diagonals from the top to the bottom) : $C_1$ for the pair $(1,2)$, $C_2$ for $(2,3)$, $\dots$, $C_{n-1}$ for $(n-1,n)$, $C_n$ for $(1,3)$, $\dots$, $C_{2n-3}$ for $(n-2,n)$,$\dots$, $C_{\frac{n(n-1)}{2}}$ for $(1,n)$.

Then, for $r=0,\dots,\frac{n(n-1)}{2}$, set $E_r=\left\lbrace (a_{i,j})\in\mathbf{U}(n,p): a_{i,j}=0\text{ for }(i,j)=C_l, l\leq r\right\rbrace$.

In such a way, one gets a central series $1=E_{\frac{n(n-1)}{2}}\trianglelefteq\dots\trianglelefteq E_1\trianglelefteq E_0=\mathbf{U}(n,p)$.

I would like to know if this series is characteristic (i.e. each term is stable by any automorphism of $\mathbf{U}(n,p)$). The problem is that I have no intuititon on the outer automorphisms of $\mathbf{U}(n,p)$: what do they look like?

Separately: I do not even know whether $\mathrm{Out}(\mathbf{U}(n,p))$ is solvable or not.

I think there is an automorphism that interchanges the $(1,2)$ and $(n-1,n)$ entries of the matrix, so $E_1$ is not characteristic. — Derek Holt, Jan 02 '20 at 17:18
@Derek Holt: I think your map is not a morphism. Yet, one may take the symmetry with respect to the anti-diagonal: this answers my question, since $E_1$ is not stable under this automorphism. — Gaussian, Jan 02 '20 at 17:43
$P$ is generated matrices ${ a_{i,i+1} : 1 \le i < n }$, where $a_{i,j}=I_n + E_{i,j}$ is the identity matrix together with a single $1$ in the $(i,j)$ position. What I meant to say was that there is an automorphism that interchanges $a_{i,i+1}$ with $a_{n-i,n-i+1}$ for each $i$. — Derek Holt, Jan 02 '20 at 18:00

Characteristic subgroups of the group of unipotent matrices

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