Say you have $E_1/K_1 \rightarrow E_2/K_2$
How would you even define such isomorphism? Assuming it is a rational map then it can be expressed as a pair of polynomials. In what field are the coefficients in? This is expressed usually when you say "a map/curve is defined over $K$" i.e. the coefficients are from $K$.
Imagine in your example $K_1 = \mathbb{F}_p, K_2 = \mathbb{F}_t$ where $gcd(t,p)=1$ then clearly $K_1$ is not a subfield of $K_2$ and vice versa. Also their algebraic closures are different.
Now say you have $Q = (a,b) \in E_1(\mathbb{F}_p)$ so $Q=(a,b)$ is a pair of elements in $\mathbb{F}_p$. $(c,d) = \phi(Q) \in E_2(\mathbb{F}_t)$ is a pair of elements in $\mathbb{F}_t$.
In more detail, the value $c$ is calculated as $c = \frac{f(a,b)}{g(a,b)}$ where $f,g$ are polynomials in 2 variables with coefficients in $K_1$ or $K_2$ (for the sake of the argument).
If $f(x,y)\in K_1[x,y]$ then $f(a,b)$ is an element of $K_1$ and also has to be of $K_2$ since we assume $c$ is in $K_2$. Therefore $K_1$ is a subfield of $K_2$.
If $f(x,y)\in K_2[x,y]$ then when calculating $f(a,b)$ you get a problem. You need to do field operations between elements of $K_1$ ($a$ and $b$) and $K_2$ (coeffs in $f$). How do you do that? How do you multiply $v \in \mathbb{F}_5$ with $w \in \mathbb{F}_7$ for example? It does not make sense. It makes sense if there is some relation between $K_1$ and $K_2$ being subfields but not in general.
I think in general, you need $K_1$ to be a subfield of $K_2$ but I am not sure at the moment. But this is just meant to ilustrate why the question does not really make sense.