Problem: If $P$ is a Sylow-$p$-subgroup of a group $G$ and $f:G\rightarrow G$ is an isomorphism then $f(P)=P$.
I've tried to make a proof through the 2nd Sylow Theorems and through the fact that $f(P)$ is a $p$-subgoup of $G$ but I think that this there is a chance that this might not be the correct approach. Any kind of help is much appreciated!
That's certainly not true -- take any group with more than one Sylow $p$-subgroup. Sylow's second theorem says all the Sylow $p$-subgroups are conjugate, so in particular there's an inner automorphism moving one to any other.
Now if you know the Sylow $p$-subgroup is unique, then it is true. Indeed, $f(P)$ is a $p$-subgroup with the same cardinality as $P$, so must be $P$.