On conjugacy of matrices in extension rings

Question

Let $p$ be any (odd) prime number and consider the group $G=\mathbb{Z}_{p^n}\times\mathbb{Z}_{p^n}$. Let $\varphi,\psi$ be automorphisms of $G$ such that the automorphisms induced on $G/pG$ are conjugate. Is it true that $\varphi$ and $\psi$ are conjugate as automorphisms of $G$?

EDIT

$\varphi$ and $\psi$ should be $p'$-automorphisms.

Answer 1

It's incredibly important that you want $p'$-automorphisms. Let $P$ be a finite $p$-group, and let $\Phi(P)$ denote the smallest normal subgroup whose quotient is elementary abelian. (In your case, this is $pG$.) If $\phi$ is a $p'$-automorphism such that $\phi$ induces the identity on $P/\Phi(P)$, then $\phi=1$.

So if one of $\phi$ and $\psi$ were the identity on $G/pG$ then you would be happy. But why don't you assume that $\phi$ and $\psi$ are equal on $G/pG$? If they are conjugate then you can do this. Then $\phi\psi^{-1}$ is the identity on $G/pG$, hence is a $p$-automorphism of $G$.

But this is all you can say. If $n\geq 2$ then the group of automorphisms is $Q\rtimes \mathrm{SL}_2(p)$, where $Q$ is some $p$-group. I can easily choose two $p'$-automorphisms lying in two different conjugates of $\mathrm{SL}_2(p)$, with the same action on the quotient, whose product lies in $Q$.

Answer 2

You can take $\varphi$ to be the identity, and say $\psi$ given by the matrix $$\pmatrix{1&p\\0&1}.$$ As this matrix is the identity modulo $p$, $\psi$ induces the identity on $G/pG$.