I think you just didn't do the calculations properly.
Let $C_n$ denote the cyclic group of order $n$ (so I don't have to type $\mathbb{Z}_5$ which is harder to $\LaTeX$ and can also be confused with the $5$-adic integers. Let $H=C_5\times C_5$.
Let $\varphi,\theta\colon\mathbb{C}_3\to\mathsf{GL}_2(5)$ be two nontrivial homomorphisms. We want to show that $H\rtimes_{\varphi}C_3\cong H\rtimes_{\theta} C_3$. This will require some specific facts about your specific orders, since the result is not true in general.
An element of order $3$ in $\mathsf{GL}_2(5)$ satisfies $X^3-1=(X-1)(X^2+X+1)$. Since both factors are irreducible in $\mathbb{F}_5$, it means that the minimal polynomial is either $X^2+X+1$ or $X-1$; in the latter case it is the identity, which we are excluding, so the minimal polynomial is $X^2+X+1$. That means that matrix for this element (with, say, the standard basis for $C_5\times C_5$) must be conjugate to the companion matrix of $X^2+X+1$. Thus, any two elements of order $3$ in $\mathsf{GL}_2(5)$ are conjugate in $\mathsf{GL}_2(5)$. That is, there exists an automorphism $\psi\colon H\to H$ such that $\theta = \psi\varphi\psi^{-1}$.
If we denote the automorphism $\varphi(y)$ by $\varphi_y$, and the automorphism $\theta(y)$ by $\theta_y$, we have that for each $y\in Y$, $\theta_y = \psi\varphi_y\psi^{-1}$.
(Note: I write my semidirect products backwards from how you write them: the first coordinate lies in the normal subgroup, the second in the "acting" group. So the product in $A\rtimes_g B$ is $(a_1,b_1)(a_2,b_2) = (a_1g_{b_1}(a_2), b_1b_2)$. You can translate this to your notation..)
Define $f\colon H\rtimes_{\varphi}C_3 \to H\rtimes_{\theta} C_3$ as follows:
$$f(x,y) = (\psi(x), y).$$
We have
$$\begin{align*}
f(x,y)f(r,s) &= (\psi(x),y)(\psi(r),s)\\
&= \Bigl(\psi(x)\theta_y\bigl(\psi(r)\bigr),ys\Bigr)\\
&= \Biggl(\psi(x)\biggl(\psi\Bigl(\varphi_y\bigl(\psi^{-1}(\psi(r)\bigr)\Bigr)\biggr),ys\Biggr)\\
&= \Bigl(\psi(x)\psi\bigl(\varphi_y(r)\bigr),ys\Bigr)\\
&= \Bigl(\psi\bigl(x\varphi_y(r)\bigr),ys\Bigr)\\
&= f\Bigl(x\varphi_y(r),ys\Bigr).
\end{align*}$$
Now note that in $H\rtimes_{\varphi}C_3$, we have
$$(x,y)(r,s) = (x\varphi_y(r),ys).$$
Therefore, we have that
$$f\Bigl((x,y)(r,s)\Bigr) = f(x,y)(r,s).$$
The map $g\colon H\rtimes_{\theta}C_3 \to H\rtimes_{\varphi}C_3$ given by $g(x,y) = (\psi^{-1}(x),y)$ is a homomorphism by the same argument, and it is clearly the inverse of $f$, so $f$ is an isomorphism, as desired.
The calculation shows that in general, if $\varphi,\theta\colon K\to \mathrm{Aut}(N)$, and there exists $\psi\in\mathrm{Aut}(N)$ such that $\varphi=\psi\theta\psi^{-1}$, then $N\rtimes_{\varphi}K\cong N\rtimes_{\theta}K$; the result is not restricted to finite groups.
