Isomorphism on a Sylow-$p$-subgroup of a group $G$. 
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!
 A: 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 $P$.
A: What is true is the following. Some terminology: an automorphism of $G$ is said to be fixed-point free, if it leaves only the identity $1$ fixed. If $f \in Aut(G)$ is fixed-point free, then $G=\{x^{-1}f(x): x \in G\}$: for $x^{-1}f(x)=y^{-1}f(y)$ is equivalent to $yx^{-1}=f(y)f(x)^{-1}=f(y)f(x^{-1})=f(yx^{-1})$, whence $yx^{-1}=1$, so $x=y$.
Theorem Let $f \in Aut(G)$ be fixed-point free, then there exists a $P \in Syl_p(G)$ such that $f(P)=P$. Moreover, there is only one such $P$.
Proof Note that $f(P)=P$ does not mean that $f$ fixes $P$ pointwise (otherwise $P$ would be trivial ...), but as a set!
Existence Since $f$ is an automorphism, $f(P) \in Syl_p(G)$. Hence by Sylow's Theorem a $y \in G$ exists with $f(P)=y^{-1}Py$. Choose $x \in G$ with $y=x^{-1}f(x)$. We claim that $f(xPx^{-1})=xPx^{-1}$: since both sets have the same cardinality, it suffices to show that $f(xPx^{-1}) \subseteq xPx^{-1}$. Take $s \in P$ and note that $f(x^{-1})=f(x)^{-1}=y^{-1}x^{-1}$. As $f(P)=y^{-1}Py$, $f(s)=y^{-1}ty$ for some $t \in P$. This amounts to: $f(xsx^{-1})=f(x) \cdot y^{-1}ty \cdot f(x)^{-1}=xy \cdot y^{-1}ty \cdot y^{-1}x^{-1}=xtx^{-1} \in xPx^{-1}$, as wanted.
Uniqueness To prove the uniqueness with respect to be invariant under $f$, assume $P \in Syl_p(G)$ being invariant under $f$. It is then is easy to see that its normalizer is also invariant under $f$, that is $f(N_G(P))=N_G(P)$. This induces an automorphism $f \mid _{N_G(P)}$, which of course is fixed-point free again, whence every element of $N_G(P)$ can be written as $n^{-1}f(n)$, with $n \in N_G(P)$.
Now assume there is another $Q \in Syl_p(G)$ with $f(Q)=Q$. Of course by Sylow's Theorem, $Q=uPu^{-1}$ for some $u \in G$. It follows that $uPu^{-1}=Q=f(Q)=f(uPu^{-1})=f(u)f(P)f(u^{-1})=f(u)Pf(u)^{-1}$, implying that $u^{-1}f(u) \in N_G(P)$. Hence $u^{-1}f(u)=n^{-1}f(n)$ for some $n \in N_G(P)$. But this implies $u=n$ and hence $Q=P$. $\square$
A: A subgroup $H\le G$ such that for any automorphism $a$ of $G$ we have $a(H)=H$ is called characteristic.
Sylow subgroups are in general not characteristic,  as we often have more than one,  which by a Sylow theorem are conjugate.   And conjugation by an element of the group is always an automorphism.
Examples of characteristic subgroups are the commutator subgroup,  the center,  the Frattini subgroup and the Fitting subgroup.
If you have only one subgroup of a certain order, then it's characteristic. For this reason,  all subgroups of cyclic groups are characteristic.
When the Sylow subgroup is unique, or normal,  it's characteristic.
