First, We set notations as follows.
$G$ : topological group , $k$ : field , $V$ : linear topological space over $k$ ,
$\mathrm{Map}(V,V)$ : Set of all continuous maps from $V$ to $V$
$\mathrm{Aut}_k (V)$ : Set of all homeomorphism from $V$ to $V$
We give compact-open topology to $\mathrm{Map}(V,V)$ and its subpace topology to $\mathrm{Aut}_k (V)$ .
Let $\rho : G \rightarrow \mathrm{Aut}_k (V)$ be a group homomorphism between topological spaces.
Then , are following conditions equivalent $???$
$(1)$ $\rho$ is a continuous map between topological spaces.
$(2)$ $G \times V \rightarrow V , (g,x) \mapsto \rho(g)(x)$ is a continuous map.
For spaces $X,Y$ let $C(X,Y)$ denote the set of continuous functions $X \to Y$. This set endowed with the compact-open will be denoted by $Y^X$. There is a canonical function $E : C(X \times Y,Z) \to C(X,Z^Y)$ where for $f : X \times Y \to Z$ we define $E(f) : X \to Z^Y$ by $E(f)(x) : Y \to Z, E(f)(x)(y) = f(x,y)$. This function is known as the exponential correspondence. See any book on general topology treating function spaces. There are also a number of contributions in this forum, for example When is the exponential law in topology discontinuous? (search for "exponential law").
The function $E$ is trivially injective, but surjectivity requires to assume that $Y$ is locally compact.
For your question this means that (2) implies (1). The converse is true under the additional assumption that $V$ is locally compact. Thus, if $k$ is a locally compact topological field, then (1) and (2) are equivalent because $V \approx k^n$ is locally compact.