The semidirect product $(C_7\times C_{13})\rtimes C_3$ Following this page, in the classification of groups of order $273$, the product of the Sylow group $S:=C_7C_{13}\simeq C_{7}\times C_{13}$ is normal, hence can be acted on by the Sylow $C_3$ to produce semidirect products.
Since $\operatorname{Aut}(S)\simeq C_6\times C_{12}$, one only needs to find an order 3 element there to produce the homomorphism $C_3\rightarrow \operatorname{Aut}(S)$.
In particular, if $C_6=\langle x\rangle$ and $C_{12}=\langle y\rangle$, up to choice of generators of $C_3$, the group is classified by the 5 actions corresponding to:
$(1,1), (x^2,1), (1,y^4), (x^2,y^4), (x^{-2},y^4)$.
It is easy to see that the center of the corresponding groups has order $273, 13, 7, 1, 1$ respectively. So my question is: how do we show that the last two groups are actually non-isomorphic?
P.S. In a similar example with $(C_7\times C_7)\rtimes C_3$ on the same page, the last two groups can be distinguished since every subgroup of order $7$ can be shown to be normal in one but not the other.
 A: Edit: I didn't realize you were talking about $x$ and $y$ being in the automorphism group, and not the group itself.  I've edited below to correct this.
We can take $\alpha$ as a generator of $C_7$, and then $x(\alpha)=\alpha^5$. So $x^2(\alpha) = \alpha^4$ and $x^{-2}(\alpha) = \alpha^2$.
We can take $\beta$ as a generator of $C_{13}$, and then $y(\beta)=\beta^2$, and so $y^4(\beta) = \beta^3$.
Consider your groups as $G=C_{91}\rtimes C_3$ and $H=C_{91}\rtimes C_3$. We can write $z$ as the generator of $C_{91}$, and $g$ and $h$ as the generator of the $C_3$ subgroups.
Then in $G$, we have $g^{-1}zg=z^{-10}$, since $-10$ is $4\pmod{7}$ and $3\pmod{13}$.
In $H$, we have $h^{-1}zh=z^{16}$, since $16$ is $2\pmod{7}$ and $3\pmod{13}$.
If $f:G\rightarrow H$ was an isomorphism, then we would have $f(z)=z^k$ for some integer $k$. We would also have $f(g)$ is some conjugate of $h$ or $h^2$, which thus acts the same as $h$ or $h^2$ on $z$.
So then
\begin{align*}
z^{-10k} &= f(z^{-10})\\
&= f(g^{-1}zg)\\
&= f(g^{-1})z^kf(g)\\
&= z^{16k}\text{ or }z^{74k}
\end{align*}
This implies $z^{26k}=1$ or $z^{84k}=1$, so that $7$  or $13$ divides $k$ respectively, which means $z^k$ is not a generator of $C_{91}$ in $H$.  This is the contradiction.
This same idea can be used whenever you have a semidirect product $A\rtimes B$, and $A$ is abelian and $B$ is cyclic. It can be used to differentiate different semidirect products coming from non-conjugate images of $B$ in $\textrm{Aut}(A)$.
