How to construct the homomorphism in semidirect product of $Z_3$ and $Z_{13}$?

I know that in the semidirect product of $$A$$ and $$B$$, the homomorphism $$\phi:A\rightarrow Aut(B)$$ should be $$\phi_y(x) = yxy^{-1}$$ but have no idea how to construct one for $$\phi:Z_3\rightarrow Aut(Z_{13})$$. Any help is appreciated. The presentation of such a group is given here Finding presentation of group of order 39 but I don't know what the explicit homomorphism would be.

• “Homeomorphism” is a topological term: it means a continuous bijection with continuous inverse. Presumably, you mean homomorphism. – Arturo Magidin Dec 3 '18 at 21:57
• See this question. – Dietrich Burde Dec 3 '18 at 22:01
• @ArturoMagidin Yes, corrected that. – manifolded Dec 3 '18 at 22:02
• ... not everywhere... but now it’s fixed. – Arturo Magidin Dec 3 '18 at 22:12
• @DietrichBurde I wanted to know which homomorphism corresponds to the presentation $\{x,y|x^{13}=y^3=1,yxy^{-1} = x^3\}$? And that question answers how to find all the homomorphisms. – manifolded Dec 3 '18 at 22:20