Let $T \in \mathcal{L}(V)$, where $\mathcal{L}(V)$ is the set of linear operators mapping a vector space $V$ to itself, and let $U$ be an isomorphism from $V$ to another vector space $W$. We claim that $T \mapsto UTU^{-1}$ determines an isomorphism from $\mathcal{L}(V)$ to $\mathcal{L}(W)$. I was wondering if this proof of this claim is valid:
Firstly, we know that $UTU^{-1}$ is linear, since a composition of homomorphisms is a homomorphism. As for whether this is an isomorphism, we know that an isomorphism is a bijective homomorphism. Now, a map $f$ is bijective if and only if it's invertible - that is to say $f^{-1}$ exists - and composition $f^{-1} \circ f$ is the identity map on the domain of $f$. So since $U$ is an isomorphism, we guess that $U^{-1}(x)U$ is the inverse of $U(x)U^{-1}$ and try to show this; sure enough, $U^{-1}(UTU^{-1})U$ can be manipulated under associativity of homomorphic composition to show that $(U^{-1}U)T(U^{-1}U) = Id(T(Id(V)) = T$.
Is this sound? I appreciate insightful commenters offering nice alternative approaches and such - and feel free to do so - but I hope to get some feedback on whether or not my specific approach given here is a valid one.
Define $f\colon \mathcal L(V)\mapsto \mathcal L(W)$ by $f(T)=UTU^{-1}$. It's easy to show that $f$ is a linear map; just verify $f(S+T)=f(S)+f(T)$ and $f(\lambda T)=\lambda f(T)$ for all $S,T\in\mathcal L(V)$ and any scalar $\lambda$. Now $$f(T)=\mathrm{id}_W\iff UTU^{-1}=\mathrm{id}_W\iff T=U^{-1}\mathrm{id}_WU=\mathrm{id}_V,$$ hence $\ker(f)=\mathrm{id}_V$, so $f$ is an isomorphism between $\mathcal L(V)$ and $\mathcal L(W)$.
Clarification: you don't have to show that $U^{-1}TU$ is a linear map (from $V$ to $W$), but that $f$ is a linear map from $\mathcal L(V)$ to $\mathcal L(W)$, that is $f\in\mathrm{Hom}\bigl(\mathcal L(V),\mathcal L(W)\bigr)$.