I have three semidirect products, two of which I believe to be isomorphic. Their presentations are
\begin{align} &\langle a, b \mid a^4 = b^{17} = 1, aba^{-1} = b^4\rangle\\ &\langle a, b \mid a^4 = b^{17} = 1, aba^{-1} = b^{13}\rangle\\ &\langle a, b \mid a^4 = b^{17} = 1, aba^{-1} = b^{16}\rangle \end{align}
I think the first of these three are isomorphic since $13^4 \equiv 4^4 \equiv 1 \bmod 17$, while $16^2 \equiv 1 \bmod 17$, so that the conjugation relations $aba^{-1} = b^4$ and $aba^{-1} = b^{13}$ can be represented equivalently by the relation $(aba^{-1})^2 = b^{16}$, while the relation $aba^{-1} = b^{16}$ gives $(aba^{-1})^2 = b$. Is this enough to conclude that the first two groups are isomorphic while the last group is distinct from the others?
Since $aba^{-1} = b^4 \Leftrightarrow a^{-1}ba = b^{-4}= b^{13}$, there is an isomorphism between the first two groups with $a \mapsto a^{-1}$, $b \mapsto b$.
To show that the groups defined by the first and third presentations are not isomorphic, you could (for example) observe that they both have unique subgroups $\langle a^2,b \rangle$ of index $2$ and that this is the dihedral group for the first presentation and an abelian group for the second. Equivalently, the third group has elements of order $34$, and the first does not.