It is known that any group is isomorphic to a group of permutations. But if $\phi: G\to T$ is an isomorphism defined as $\phi_g(x) =gx$ for elements $g, x \in G$, then how can it be that $T_{g^{-1}}=[T(g)]^{-1}$? Because $T_{g^{-1}}(x)=g^{-1}x$, but $[T(g)]^{-1}=(gx)^{-1}=x^{-1}g^{-1}\neq g^{-1}x$.
What am I missing?
On
What you are missing is the meaning of $[T(g)]^{-1}$. Indeed in what you wrote you say that :
$$[T(g)]^{-1}\text{ is the function from }G\text{ to } G \text{ sending } x\in G \text{ to } [T(g)(x)]^{-1}=(gx)^{-1}=x^{-1}g^{-1} $$
But $[T(g)]^{-1}$ does not mean this, it is the inverse mapping of $T(g)$. By definition (if it exists) it sends $T(g)(x)$ on $x$ for any $x\in G$.
Now $T(g)^{-1}(T(g)(y))=y$ so that $T(g)^{-1}(g.y)=y$ this is true for any $y\in G$, in particular if $y=g^{-1}.x$ :
$$g^{-1}.x=T(g)^{-1}(g.(g^{-1}.x))=T(g)^{-1}((gg^{-1}).x)=T(g)^{-1}(x)$$
If it exists then $T(g)^{-1}(x)=g^{-1}.x$. There is no choice. One can see that this is indeed the inverse function for $T(g)$.
Claim: If $\phi: G \to T$ is an isomorphism, then $\phi(1_G) = 1_T$.
For this reason, left translation (or left multiplication) does not give an isomorphism except when $g = 1$, in which case $\phi_1 = \operatorname{id}_G$, the identity map on $G$.