Is there a way to describe the structure of $Aut(UT(3, p))$? > Is there a way to describe the structure of the automorphism group of $$C_{p}^2 \rtimes C_p \cong \langle x, y, z | [x,y]=z, [x,z]=[y,z]=x^p=y^p=z^p=e \rangle \cong UT(3, p)?$$ Here $p$ is an odd prime. The only thing I know about it is, that $Inn(UT(3, p)) \cong \frac{UT(3, p)}{Z(UT(3, p))} \cong C_p \times C_p$, however $UT(3, p)$ is also very likely to have outer automorphisms, which I do not know how to describe. Also, one can see, that all inner automorphisms of $UT(3, p)$ are of the form $$\begin{pmatrix} 1 & x & y \\\ 0 & 1 & z\\\ 0 & 0 & 1 \end{pmatrix} \mapsto \begin{pmatrix} 1 & x & y + (a-c)z -ac\\\ 0 & 1 & z\\\ 0 & 0 & 1 \end{pmatrix}$$ for some $a, c \in \mathbb{F}_p$. However, this does not help much.

For $p\geq3$ we have $Out(G)=GL_2(p)$ and $Aut(G)=AGL_2(p)\cong C_p^2\rtimes GL_2(p)$. To see this it helps to think of the group as $\mathbb{F}_p^2\rtimes C_p$ with the $C_p$ factor acting linearly, then the outer automorphisms come from $GL_2(p)$ acting by conjugation. Equivalently, the automorphisms essentially permute the subgroups of order $p^2$ containing the center, while leaving the center itself unmoved.

This) may be useful as well.

\--Edit--

As pointed out by Derek Holt in the comments, viewing the group as the above semidirect product is not the clearest way to see the automorphism group. It is better to think of $Out(G)$ as a sympletic group preserving the alternating form defined by commutators in $G$. This will still give $Out(G)\cong GL_2(p)$ and $Aut(G)\cong AGL_2(p)$.