The answer to your question is no: actually you can have $\varphi_1$ and $\varphi_2$ both injective (so with the same kernel) and still have non-isomorphic semi-direct products.
For instance, take $K=\mathbb{Z}/2\mathbb{Z}$ and $H=\mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$, and define $\varphi_1$ such that the non-trivial element in $K$ acts non-trivially on the $\mathbb{Z}/3\mathbb{Z}$ factor (and trivially on the other one), and $\varphi_2$ such that it acts non-trivially on $\mathbb{Z}/4\mathbb{Z}$ (and trivially on the other one).
Then $\varphi_1$ and $\varphi_2$ are injective, but the semi-direct products are respectively $D_6\times \mathbb{Z}/4\mathbb{Z}$ and $D_8\times \mathbb{Z}/3\mathbb{Z}$ (where $D_{2n}$ is the dihedral group with $2n$ elements) which are not isomorphic (for instance we can look at the center).
Here's an example that might help: Let $X$ have order $7$ and $Y$ have order $3$. These groups are cyclic; write $x$ for a generator of $X$ and $y$ for a generator of $Y$. Let $\phi\colon X \to X$ and $\psi\colon X \to X$ be the automorphisms $\phi(x) = x^2$, and $\psi(x) = x^4$.
Notice that $\phi^3(x) = x^{2^3} = x^8 = x$, and $\psi^3(x) = x^{4^3} = x^{64} = x$, so as automorphisms of $X$, $\phi$ and $\psi$ each have order three.
I claim that $X\rtimes_\phi Y \cong X\rtimes_\psi Y$. Indeed, (abusing notation slightly) $X\rtimes_\phi Y$ and $X\rtimes_\psi Y$ are generated by $x$ and $y$, so to specify a homomorphism, I just need to tell you where $x$ and $y$ go, and then check that the map I've written down indeed defines a homomorphism. I claim that the map $x \mapsto x$, $y \mapsto y^{-1}$ is such a homomorphism and moreover that it is an isomorphism. I'll leave it to you to check this.
So if the answer has to be "sometimes" and not "never," maybe we should reconsider what should be true based on this example. Aut$(X)$ is generated by $x \mapsto x^3$. There is an automorphism of Aut$(X)$ that sends the automorphism $x \mapsto x^3$ to the automorphism $x \mapsto x^5$. Under this automorphism of Aut$(X)$, $\phi$ is mapped to $\psi$.
So some questions to investigate: if we have $\phi$ and $\psi$ in Aut$(X)$, will it be the case that $X\rtimes_\phi Y\cong X\rtimes_\psi Y$ whenever there is some automorphism $f\colon$ Aut$(X) \to $ Aut$(X)$ such that $f(\phi) = \psi$? Is this the only condition?
Best Answer
Suppose $G$ is a $p$-group and $H$ is a $q$-group for different primes $p,q$.
Then $G$ is a characteristic subgroup of $G \rtimes H$ and $H$ is specified up to conjugacy as a Sylow $q$-subgroup.
In particular, for any two semi-direct products $X_i = G \rtimes_{\phi_i} H$ and any isomorphism $f:X_1 \to X_2$ we are guaranteed that $f(G)=G$ and $f(H)$ is $X_2$-conjugate to $H$. In other words, there is an inner automorphism $\nu$ of $X_2$ such that the composition $X_1 \xrightarrow{f} X_2 \xrightarrow{\nu} X_2 = X_1 \xrightarrow{\bar f} X_2$ has the property that $\bar f(G) = G$ and $\bar f(H) = H$.
Consider the centralizer $C_i = C_H(G)$ inside the two groups $X_i$. $$\begin{array}{rl} C_i &= \{ (1,h) : (1,h)\cdot_i (g,1) = (g,1) \cdot_i (1,h) \forall g \in G \} \\ &= \{ (1,h) : (\phi_i(h)(g),h) = (g,h) \forall g \in G \} \\ &= \{ (1,h) : \phi_i(h)(g) = g \forall g \in G \} \\ &= \{ (1,h) : \phi_i(h) = 1 \} \\ &= \{ (1,h) : h \in \ker(\phi_i) \} \\ \end{array} $$
In other words, $C_i$ is the kernel of $\phi_i$. Now $\ker(\phi_1) \neq \ker(\phi_2)$ is not particular relevant. What we care about is whether $\bar f(C_1) = C_2$ (which it must, if $f$ really is an isomorphism). However, $\bar f$ restricted to $H$ is easily seen to be an automorphism of $H$, so what matters is whether there is an automorphism of $H$ that takes $\ker(\phi_1)$ to $\ker(\phi_2)$. If there is not, then there can be no $\bar f$.
On page 187 of Dummit–Foote, exercise 7c, all of these properties are satisfied.
Technical conditions: We needed $G$ characteristic in $X_i$ and $H^1(H,G)=0$. If $G$ is a Hall $\pi$-subgroup of $X$ (for instance a Sylow), then Schur–Zassenhaus guarantees the relevant properties are satisfied. If $G$ is abelian, then we don't need the cohomology condition, just the characteristic.
Counterexample
Without the technical conditions we can have counterexamples. For example let $G=H$ be nonabelian of order 6. Let $\phi_1(h)(g) = g$ and $\phi_2(h)(g) = hgh^{-1}$. Then $\ker(\phi_1) = H$ and $\ker(\phi_2) = 1$ but $X_1 \cong X_2$. In this case $G$ is not characteristic, but we can achieve the isomorphism with an $f$ such that $f(G)=G$. In this case however, there is no $\bar f$ with $\bar f(G) = G$ and $\bar f(H) = H$.
In particular, we have non-isomorphic kernels with isomorphic semi-direct products.