Elements of order $3$ in $\text{Aut}\left(\mathbb{Z}/91\mathbb{Z}\right)$ It looks like someone has already been here, but the question I have goes farther.
To summarize my work, as well as the work in the above post, we know that $\text{Aut}\left(\mathbb{Z}/91\mathbb{Z}\right) \cong \mathcal{U}_{91} \cong \mathcal{U}_7 \times \mathcal{U}_{13}$  by the Chinese Remainder Theorem. And $\mathcal{U}_7 \times \mathcal{U}_{13} \cong \mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/12\mathbb{Z}$.
For a while I was stuck where the other post happened to leave off: I found the eight elements of order $3$ in terms of $\mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/12\mathbb{Z}$; great, but the question asks for automorphisms.  I know that any $\varphi \in \text{Aut}\left(\mathbb{Z}/91\mathbb{Z}\right)$ must map a generator to a generator, so I assumed that $\varphi(1) = k$, where $(91, k) = 1$.
If $|\varphi| = 3$, then $\varphi^3(1) = k^3 = 1$  i.e., $k^3 - 1 \equiv 0 \;(\text{mod } 91)$.  Factoring $k^3 - 1$ as $(k-1)(k^2+k+1)$, I found that $k = 9$ and $k = 22$ gave solutions. So if $\pi(1) = 9$ and $\rho(1) = 22$, then $\pi, \rho \in \text{Aut}\left(\mathbb{Z}/91\mathbb{Z}\right)$, $|\pi| = |\rho| = 3$, and the elements of order $3$ are $\left< \pi, \rho \right> \backslash \{\text{id}\}$.
I was pretty happy with this, but then I started playing around with my answers, making sure they didn't lead to nonsense. The first problem I found was that, although $|26| = 7$ in $\mathbb{Z}/91\mathbb{Z} $, as defined above $\rho(26)=26\cdot\rho(1)=26\cdot 22 = 572 \equiv 4$, and $|4| = 91$. I tried redefining $\rho$ so that $\rho(1) = 29$ (as originally defined, this would have been $\rho^2(1)$, which of course also has order $3$), and that fixes my problem with $26$.  But there are a few other numbers to check, and I'm not sure that checking them is a viable option.
So are $\pi$ and the new $\rho$ the automorphisms I want?  If so, how can I be sure?  since my original $\rho$ broke down.  If not, how do I find them, and, again, how do I know once I have?
I'll add that I suspect this may involve generators of $\mathcal{U}_7$ and of $\mathcal{U}_{13}$: $\mathcal{U}_7$ is generated by $3$, and $3^2 = 9$  hence $\pi$ above.  As for taking the fourth root of $29$ in $\mathcal{U}_{13}$ ... Even looking for a square root was taking too long.
 A: You've essentially solved the problem by finding the elements of order $3$ in $\mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/12\mathbb{Z}$, but now you need to chase back through the various isomorphisms to find the corresponding automorphisms.
You're right, to realize the isomorphism $(\mathbb{Z}/7\mathbb{Z})^\times \times (\mathbb{Z}/13\mathbb{Z})^\times \cong \mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/12\mathbb{Z}$, we need to find generators for $(\mathbb{Z}/7\mathbb{Z})^\times$ and $(\mathbb{Z}/13\mathbb{Z})^\times$.  You found that $3$ generates $(\mathbb{Z}/7\mathbb{Z})^\times$ and since
\begin{align*}
2^4 &\equiv 16 \equiv 3 \pmod{13}\\
2^6 &\equiv 3 \cdot 4 \equiv 12 \pmod{13}
\end{align*}
then $2$ generates $(\mathbb{Z}/13\mathbb{Z})^\times$.  This gives us the following isomorphism.
\begin{align*}
\mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/12\mathbb{Z} &\overset{\sim}{\to} (\mathbb{Z}/7\mathbb{Z})^\times \times (\mathbb{Z}/13\mathbb{Z})^\times\\
(a,b) &\mapsto (3^a, 2^b)
\end{align*}
Next we need to find the isomorphism $(\mathbb{Z}/7\mathbb{Z})^\times \times (\mathbb{Z}/13\mathbb{Z})^\times \overset{\sim}{\longrightarrow} (\mathbb{Z}/91\mathbb{Z})^\times$.  Since $1 = 14 - 13 = 2 \cdot 7 + (-1) 13$, then we have an isomorphism
\begin{align*}
(\mathbb{Z}/7\mathbb{Z})^\times \times (\mathbb{Z}/13\mathbb{Z})^\times &\overset{\sim}{\longrightarrow} (\mathbb{Z}/91\mathbb{Z})^\times\\
(r, s) &\longmapsto 14s - 13r \, .
\end{align*}
(See here for more on this isomorphism.)
Lastly, we have an isomorphism
\begin{align*}
(\mathbb{Z}/91\mathbb{Z})^\times &\overset{\sim}{\to} \text{Aut}(\mathbb{Z}/91\mathbb{Z})\\
k &\mapsto \varphi_k
\end{align*}
where $\varphi_k: \mathbb{Z}/91\mathbb{Z} \to \mathbb{Z}/91\mathbb{Z}$ is given by $\varphi_k(x) = kx$.
Taking the elements of order $3$ you found in $\mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/12\mathbb{Z}$ and using these isomorphisms, you should be able to find the automorphisms of $\mathbb{Z}/91\mathbb{Z}$ of order $3$.
A: Note SpamIAm's method yields:
$(2,0) \mapsto \phi_{79}$ and $(0,4) \mapsto \phi_{29}$.
Explicitly calculating $\langle \phi_{79},\phi_{29}\rangle$, we find the subgroup of $\text{Aut}(\Bbb Z_{91})$ isomorphic to $\Bbb Z_3 \times \Bbb Z_3$ is:
$\{\phi_1,\phi_{79},\phi_{53},\phi_{29},\phi_{22},\phi_{16},\phi_9,\phi_{81},\phi_{74}\}$
So your $\phi_9$ and $\phi_{22}$ are indeed generators-your error was that $572 = 26$ (mod $91$), and clearly $26$ has order $\dfrac{91}{\gcd(26,91)} = 7$.
