Let $G$ be a finite group and $\gamma \in \text{Sym}(G)$ such that $\gamma (1) = 1$ and $\gamma (gH) = \gamma (g)H$ for all $g\in G$ and $H\leq G$.
This means $\gamma$ induces a permutation of the left cosets of any subgroup of $G$.
I need to show that $\gamma$ fixes any left coset of the Center $C(G)$ of $G$, that is $$\gamma(g)C(G) = gC(G)\,.$$
I tried some simple calculations which did not work. Now I think some more structural arguments are necessary. Any ideas?
I found a counterexample using $\textsf{GAP}$. The group is $$ G \;=\; \mathbb{Z}_9 \rtimes \mathbb{Z}_3 \;=\; \langle a,b \mid a^9=b^3=1,b^{-1}ab=a^4\rangle, $$ with the function $\gamma\colon G\to G$ defined by $$ \gamma\bigl(a^j b^k\bigr) \;=\; a^{-j} b^{-k}. $$ According to $\textsf{GAP}$, this permutes the left cosets of each subgroup of $G$. Indeed, the proper, nontrivial subgroups of $G$ are
The cyclic subgroups $\langle a^3\rangle$, $\langle b\rangle$, $\langle a^3b\rangle$, and $\langle a^6b\rangle$ of order $3$,
The cyclic subgroups $\langle a\rangle$, $\langle ab\rangle$, and $\langle ab^2\rangle$ of order $9$, and
The subgroup $\langle a^3,b\rangle$ of order $9$,
so one could check this statement by hand without too much trouble. However, the center of $G$ is $\langle a^3\rangle = \{1,a^{-3},a^3\}$, so $\gamma$ switches the cosets $\{a,a^{-2},a^4\}$ and $\{a^{-1},a^{-4},a^2\}$.
Here are a few more observations and comments:
The statement is true if $\gamma$ is an automorphism of $G$. For then $\gamma$ must fix every subgroup of $G$, so it is a power automorphism. By a result of C. Cooper, it follows that $\gamma$ is central in the automorphism group, so by the argument at the end of Matt Samuel's (now deleted) post, it follows that $\gamma$ fixes all cosets of the center.
A bijection $\gamma$ between two groups satisfying the given conditions is known as a (normed) $\mathfrak{R}$-isomorphism. See Section 9.4 of Schmidt, Subgroup Lattices of Groups. (The part on $\mathfrak{R}$-isomorphisms starts on pg. 530.)
