Let $G=\langle a,b,c : a^9=b^3=1,c^3=a^3,ab=ba^4,ac=cab^{-1},cb=bc\rangle$.
Find $L(G)$ where $L(G)=\{x\in G:\alpha(x)=x \; \forall\alpha\in\text{Aut}(G)\}$
By using GAP, the following are the information I have:
(i) $G$ is of size $81$.
(ii)$Z(G)=\langle a^3\rangle$.
(iii) $G$ is nilpotent group of class $3$.
(iv) $|\text{Aut}(G)|=486$
By using the commutator notation, $G$ can be also written as $$\langle a,b,c : a^9=b^3=1,c^3=a^3,[a,b]=a^3,[c,a]=b,[b,c]=1\rangle$$
It can be easily checked that $L(G)\leq Z(G)$. Hence $L(G)=1$ or $\langle a^3 \rangle$. So the problem is solved if one can find an automorphism that does not fix $a^3$ or show that $a^3$ is fixed under every automorphism.
If I want to show the latter part, let $\alpha\in \text{Aut}(G)$, then $\alpha(a)=a^ib^jc^k$. Hence $\alpha(a^3)=(a^ib^jc^k)^3$. Here the commutator identities cannot be used since $G$ is nilpotent of class $3$. So I can't find the restriction on the values of $i,j,k$.
Any hints or idea would be much appreciated. Thanks.
Writing $z=a^3=c^3$, the map $\phi$ with $a \mapsto a^2$, $b \mapsto bz$, $c \mapsto c^2z$, $z \mapsto z^2$ extends to an automorphism of $G$.
I found this automorphism by computer, but it is not difficult to verify by hand that it preserves the group relations, and hence extends to an automorphism.
For example, consider the relation $c^{-1}ac=ab^{-1} = ab^2$. Then $\phi(a)\phi(b^2) = a^2b^2 z^2$, whereas $\phi(c^{-1})\phi(a)\phi(c) = z^{-1}c^{-2}a^2c^2z=c^{-2}a^2c^2$.
Now $c^{-2}ac^2 = ab^2b^2= ab$, so $c^{-2}a^2c^2=abab=a^2b^2z^2$, because $ba=abz^2$.