A finite 2-group containing the dihedral group of order 16?

Question

The dihedral group $D_{16}$ of order 16 has a presentation $D_{16}= \langle a,t \ | \ a^2=t^8=atat=e\rangle$.

Question: Does there exist a finite 2-group $G$ containing $D_{16}$ as a subgroup, and an element $g \in G$ such that $gag^{-1}=t^4$? Bonus pats-on-the-back if $G$ has order 64.

An obvious reduction: one can assume that $G =\langle D_{16},g\rangle$.

An obvious constraint: $D_{16}$ cannot be normal in $G$ (so $G$ can't have order 32).

[This question has come up in investigations of the Balmer spectrum of $G$-equivariant stable homotopy for finite $p$-groups $G$. Like Dr. Frankenstein, I am looking for interesting subjects to experiment on, and my student Chris Lloyd is serving as the able assistant to the mad scientist.]

Answer 1

No. We can prove this by induction. Let $G$ be the smallest $2$-group in which this situation occurs. Then $G$ has a normal central subgroup $N$ of order $2$.

If $N$ has trivial intersection with the subgroup $\langle a,t \rangle = D_{16}$, then the same situation occurs in $G/N$, contradicting the minimality of $G$.

So that intersection must be nontrivial, and hence $N \le \langle a,t \rangle$, and then we must have $N = Z(\langle a,t \rangle) = \langle t^4 \rangle$.

But then $t^4 \in Z(G)$, contradicting the assumption that it is conjugate in $G$ to $a$.

The situation you describe can occur in finite groups, such as in simple groups ${\rm PSL}(2,q)$ for prime powers $q$ with $q \equiv 15$ or $17 \bmod 32$, ($q=17$ for example), but not in finite $2$-groups.