Let $dim(V)<\infty$. Let $G:=\{ f\in L(V,V) \mid f^{-1}=f^{ad}\}$ and $S(V)=\{ f\in L(V,V) \mid f\, \text{is bijective}\}$. I want to show that $G$ is a subgroup of $S(V)$.
Thus we have to show:
$\bullet \text{G is not empty}\\ \bullet \text{for all f and g in S(V): f,g in G }\to\text{ f+g in G }\\ \bullet \text{for all f in S(V) also }f^{-1} \in G $
$\bullet$ G is not empty since $F=Id_V$ is bijective and $Id_V^{-1}=Id_V=Id_v^{ad}. $ Thus $F \in G$
EDIT
To the second bullet: $(f\circ g)^{ad}=g^{ad}\circ f^{ad}=g^{-1}\circ f^{-1}=(f\circ g)^{-1}$
To the third bullet: $Id_V=(f\circ f^{-1})=(f\circ f^{-1})^{ad}=f^{{ad}^{-1}}\circ f^{ad}=f^{{-1}^{-1}}\circ f^{-1}=f\circ f^{-1}=(f^{-1}\circ f)$ Thus $f$ has an inverse element and therefore is a subgroup of G.
P.S. I have mistaken subvectore space with subgroup. But this should hold now.
The group operation in $S(V)$ is composition, and it is not closed under addition (i.e. $id, -id$ are invertible but their sum is not), or multiplication by $0$.
And, yes, a fixed basis (or an inner product) in $V$ is assumed to be able to interpret $f^{ad} $.