We define $\phi_{(1\,2\,4\,3)}$ as $\phi_{(1\,2\,4\,3)}= (1\,2\,4\,3)x(3\,4\,2\,1)$ where $x\in \text{Inn}(S_5)$ and $\text{Inn}(S_5)=\{\phi_s: s\in S_5\}$. We are interested in finding when $(\phi_{(1\,2\,4\,3)})^n=e$.
In other words when $$(1\,2\,4\,3)x(3\,4\,2\,1)(1\,2\,4\,3)x(3\,4\,2\,1)\cdots (1\,2\,4\,3)x(3\,4\,2\,1)=e.$$ This simplifies to $x^n=e$. I'm not quite sure what to do from this point on. Clearly the identity element of $\text{Inn}(S_5)$ is just $exe=x$. But this doesn't make sense as that would imply that $e^n=e$ and there is no where to go from this.
What am I missing?
You have a surjective homomorphism $\Phi: S_n \to Inn(S_n)$, $\Phi(\sigma)=\phi_{\sigma}$. As pointed out in the comments, $$\phi_{\sigma}^n=\Phi(\sigma)^n = \Phi(\sigma^n)=\phi_{\sigma^n}.$$
Now, a general property of homomorphism implies that $|\phi_\sigma|$ divides $|\sigma|$ (in your example, $|(1243)|=4$). But in this case, we also know $\ker\Phi=Z(S_n)$. What does that tell you about $\Phi$ and $|\phi_\sigma|$?