Semi-direct product of groups with one of them cyclic 
Let $K$ be a cyclic group. Let $\phi,\psi: K\rightarrow Aut(H)$ be group homomorphisms such that there exists $\zeta\in Aut(H)$ satisfying $\phi(K)=\zeta \psi(K)\zeta^{-1}$. Then can we prove $H\rtimes_{\phi}K\simeq H\rtimes_{\psi}H$?

My idea is to use the following
Theorem: Let $K,H$ be groups. Let $\phi,\psi: K\rightarrow Aut(H)$ be group homomorphisms and $\zeta\in Aut(H)$, $\alpha\in Aut(K)$ and $\psi=\sigma_{\zeta}\circ\phi\circ\alpha$, where $\sigma_{\zeta}\circ \phi\circ\alpha(k)=\zeta(\phi(\alpha(k)))\zeta^{-1}$ for all $k\in K$. Then $H\rtimes_{\phi}K\simeq H\rtimes_{\psi}H$.
Let $K=\langle a\rangle$ generated by $a$. From the condition $\phi(K)=\zeta \psi(K)\zeta^{-1}$, we have $\phi(a)=\zeta \psi(a^n)\zeta^{-1}$ for some integer $n$. Then how to continue? Thanks.
 A: The problem is that the action of $\zeta$ might not lift to the whole of $K$.
As an example, let $H=\langle z \mid z^7=1 \rangle = C_7$ and $K = \langle x,y \mid x^3=y^7=1,x^{-1}yx=y^2 \rangle$ be the nonabelian group of order $21$.
Define $\phi,\psi$ by $\phi(x)= (z \mapsto z^2), \phi(y)= 1$, and
$\psi(x)= (z \mapsto z^4), \psi(y)= 1$.
So $\phi(K)=\psi(K)$ and we can take $\zeta=1$.
The two semidirect products are
$\langle x,y,z \mid x^3=y^7=z^7=1,yz=zy,x^{-1}yx=y^2,x^{-1}zx=z^2 \rangle$, and
$\langle x,y,z \mid x^3=y^7=z^7=1,yz=zy,x^{-1}yx=y^2,x^{-1}zx=z^4 \rangle$,
which are not isomorphic.
Added later: I realize that this is not a counterexample to the question asked, because $K$ is not cyclic. But I think the example is interesting, so I will not delete it.
A: If $$\zeta \varphi(a)\zeta^{-1} = \psi(a)^t$$ I believe the map
$$
(h, a^i) \mapsto (h^{\zeta}, a^{it} ) 
$$
provides the required isomorphism from the semidirect product with respect to $\varphi$ to the one with respect to $\psi$. Here I am writing morphisms as exponents. 
(Here I am only addressing the first question, with $K = \langle a \rangle $ cyclic.) 
Addendum Having seen the second answer by Derek Holt, I now realise this is only a morphism, which should be an isomorphism only when $a \mapsto a^t$ is. 
A: Second attempt. I believe this is false when $K$ is infinite cyclic, because automorphisms of finite quotients of $K$ do not usually lift to automorphisms of $K$. For example, the two groups
$\langle x,y \mid x^{11}=1, y^{-1}xy=x^4 \rangle$ and
$\langle x,y \mid x^{11}=1, y^{-1}xy=x^5 \rangle$
(with $H = \langle x \rangle$ and $K = \langle y \rangle$) are not isomorphic, but the images of $\phi$ and $\psi$ are both equal to subgroup of ${\rm Aut}(H)$ of order $5$.
A: I think, this is false. Take $H=Z^5$, $K$ cyclic of order 5 generated by $k$, $\phi(k)$ an element $c$ of order 5 and $\psi(k)=c^2$. I think, the corresponding semidirect products are not isomorphic since $c$ is not conjugate to $c^2$ in the group $GL(5,Z)$.  
