Suppose that $\rho : K \to K$ is a group automorphism of $K$, and $\psi : K \to \operatorname{Aut}(H)$. Show that $H \rtimes_\psi K \cong H \rtimes_{\phi} K$ where $\phi = \psi \circ \rho$.
Just a helpful hint to get me started with this proof would be helpful. Not looking for the whole proof just some clues/hints/tips.
Thank you!
If we are going to use the definitions $$(a,b)\cdot_1(c,d)=(a\psi(b)c,bd),$$ and $$(a,b)\cdot_2(c,d)=(a(\psi\circ\rho)(b)c,bd),$$ then the checking of group's axioms for $H\rtimes_{\psi}K$ and $H\rtimes_{\phi}K$ are an easy exercise, where we had abbreviated $\phi=\psi\circ\rho$.
Now define $$f:H\rtimes_{\psi}K\to H\rtimes_{\phi}K,$$ via $$f(a,b)=(a,\rho(b)).$$
Now it should be clear to check that $$f((a,b)\cdot_1(c,d))=f(a\psi(b)c,db),$$ $$\qquad\qquad\qquad\qquad=(a\psi(b)c,\rho(bd)),$$ will be equal to $$f(a,b)\cdot_2 f(c,d)=(a,\rho(b))\cdot_2(c,\rho(d)),$$ $$\qquad\qquad\qquad\qquad=(a\phi(\rho(b))c,\rho(c)\rho(d)).$$
This would implies that $f$ is a group's homomorphism.
Also, the checking of bijective property for $f$ gives no big problem, I think.