I believe this is the answer in the split case: Let $E$ be the Tate curve with parameter $q$. Let $n>1$. We look for isogenies with cyclic kernel of order $n$. First, there is the isogeny to the Tate curve $E'$ with parameter $q' = q^n$ and the map is induced from $K\to K$ sending $x$ to $x^n$. With respect to the basis of $\ell^{n}$ torsion where the first element is an $\ell^n$-th root of unity $\zeta_{\ell^n}$ and the second element is a choice of an $\ell^n$-th root of $q$ (and correpondingly of $q'$), the matrix for this isogeny on $V_{\ell}(E) \to V_{\ell}(E')$ is diagonal with entries $n$, $1$.

All other cyclic isogenies of degree $n$ leaving $E$ have the kernel generated by an $n$-th root $q'$ of $q$ (and the isogeny is only defined over $K$ if this $q'$ belongs to it). The corresponding map $K/q^{\mathbb{Z}}\to K/(q')^{\mathbb{Z}}$ is induced by the identity map. This time the matrix is also diagonal by with diagonal entries first $1$ then $n$.

I believe this is the answer in the split case: Let $E$ be the Tate curve with parameter $q$. Let $n>1$. We look for isogenies with cyclic kernel of order $n$. We may suppose that $n$ is prime.

First, there is the isogeny to the Tate curve $E'$ with parameter $q' = q^n$ and the map is induced from $K\to K$ sending $x$ to $x^n$. With respect to the basis of $\ell^{n}$ torsion where the first element is an $\ell^n$-th root of unity $\zeta_{\ell^n}$ and the second element is a choice of an $\ell^n$-th root of $q$ (and correpondingly of $q'$), the matrix for this isogeny on $V_{\ell}(E) \to V_{\ell}(E')$ is diagonal with entries $n$, $1$.

All other cyclic isogenies of degree $n$ leaving $E$ have the kernel generated by an $n$-th root $q'$ of $q$ (and the isogeny is only defined over $K$ if this $q'$ belongs to it). The corresponding map $K/q^{\mathbb{Z}}\to K/(q')^{\mathbb{Z}}$ is induced by the identity map. This time the matrix is also diagonal by with diagonal entries first $1$ then $n$.