Does $\alpha=\beta f \Rightarrow f$ isomorphism?

Question

For a positive integer $n$, let be:

• $K$ and $H$ finite groups of order $n$;
• $S_n$ the symmetric group of degree $n$;
• $\alpha\colon K \hookrightarrow S_n$ and $\beta\colon H \hookrightarrow S_n$ embeddings;
• $f\colon K \rightarrow H$ bijection.

Does $\alpha=\beta f \Rightarrow f$ isomorphism? If not in general, is that true for some conditions on $\alpha$ and $\beta$?

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Edit based on @Matthias Klupsch's hint:

$(\beta f)(xy)=\beta(f(xy))$; but $\beta f$ and $\beta$ are, in particular, homomorphisms, so: $(\beta f)(xy)=((\beta f)(x))((\beta f)(y))=(\beta(f(x))(\beta(f(y))=\beta(f(x)f(y))$; therefore, $\beta(f(xy))=\beta(f(x)f(y))$; but $\beta$ is injective, so $f(xy)=f(x)f(y)$, and $f$ is homomorphism and hence isomorphism.

Answer 1

You have shown that given groups $G,K,H$ with embeddings $\alpha: K\to G$ and $\beta : H\to G$, any bijection $f:K\to H$ satisfying $\alpha = \beta f$ is necessarily an isomorphism. We don't need to impose any further restrictions.

Answer 2

Here is a simple proof that $f$ is a homomorphism.

We need to prove that $f(xy)=f(x)f(y)$. Since $\beta$ is injective, it suffices to prove that $\beta(f(xy))=\beta(f(x)f(y))$. This is the key idea.

Indeed,

$\quad \beta(f(xy)) = \alpha(xy) = \alpha(x) \alpha(y)$, because $\alpha$ is a homomorphism

$\quad \beta(f(x)f(y)) = \beta(f(x)) \beta(f(y)) = \alpha(x) \alpha(y)$, because $\beta$ is a homomorphism

and we're done.