The claim is that loop algebras $\mathcal{L}(\mathfrak{g}, \sigma)$ and $\mathcal{L}(\mathfrak{g})$ of a Lie algebra $\mathfrak{g}$ are isomorphic if the twist $\sigma$ is inner. As far as I know, Kac proved this result, but I'd like to re-construct the proof on my own.
Let's start with defining loop algebra as the space of smooth functions $\gamma(\theta) : S^1\rightarrow \mathfrak{g}$ with periodicity $\gamma(0) = \gamma(2\pi)$. If there's an inner twist, endpoints are shifted by $\sigma$, i.e $\gamma(0) = \sigma(\gamma(2\pi))$.
For a Lie group $G$ and its Lie algebra $\mathcal{g}$, adjoint action of $G$ on $\mathfrak{g}$ is given by
$$\text{Ad}_g : \mathfrak{g} \to \mathfrak{g}, \quad \text{Ad}_g =\frac{d}{dt}\biggr|_{t = 0} g \exp(tX) g^{-1}\tag{1}$$
which is a Lie algebra homomorphism that satisfies $$\mathrm{Ad}_{gh} = \mathrm{Ad}_g \circ \mathrm{Ad}_h\tag{2}$$
$$\mathrm{Ad}_g([X,Y]) = [\,\mathrm{Ad}_g X, \mathrm{Ad}_g Y\,]\tag{3}$$
Differentating this map at identity
$$\text{ad}_{X} : Y\mapsto \frac{d}{dt}\biggr|_{t = 0} \text{Ad}_{\exp(tX)}Y\tag{4}$$
gives us an inner Lie algebra derivation for $\mathfrak{g} = T_{\epsilon}G$.
Inner automorphism group of $\mathfrak{g}$ is generated by the exponential of adjoint Lie algebra derivation in $(4)$
$$\text{Inn}(\mathfrak{g}) = \langle \exp(\text{ad}_{X}) : X\in \mathfrak{g}\rangle$$
If $G$ is simply-connected, from the fundamental identity that $\text{Ad}_{\exp(X)} = \exp(\text{ad}_{X})$, we can realize $\text{Ad}(G) = \text{Inn}(\mathfrak{g})$. Furthermore, there exists a smooth path $U : [0, 2\pi]\rightarrow G$ such that $U(0) = e$ and $U(2\pi) = \exp(X)$.
Finally, let's define a Lie algebra isomorphism $$\Phi : \gamma(\theta)\in \mathcal{L}(\mathfrak{g}, \sigma)\mapsto \text{Ad}_{U^{-1}(\theta)}\gamma(\theta)$$
Where the $\mathfrak{g}$-loop takes up twist $\gamma(0) = \exp(\text{ad}_X)\gamma(2\pi)$. This is demonstrably $2\pi$-periodic $\mathfrak{g}$-valued loop:
$$\text{Ad}_{U^{-1}(0)}\gamma(0) = \text{Ad}_{e}\gamma(0) = \gamma(0)$$
$$\text{Ad}_{U^{-1}(2\pi)}\gamma(2\pi) = (\text{Ad}_{\underbrace{U^{-1}(2\pi)}_{ = \exp(-X)}}\circ\exp(\text{ad}_{X}))\gamma(0) = (\text{Ad}_{\exp({-X)}}\circ\text{Ad}_{\exp{(X)}})\gamma(0) = \text{Ad}_{e}\gamma(0) = \gamma(0)$$
Lie algebra homomorphism
For every smooth loop $\gamma_1, \gamma_2 : S^{1}\rightarrow \mathfrak{g}$, using the fact $(3)$ leads us to
$$ [\Phi(\gamma_1),\Phi(\gamma_2)](\theta) = [\mathrm{Ad}_{U^{-1}( \theta)}\gamma_1(t\theta), \mathrm{Ad}_{U^{-1}(\theta)}\gamma_2(\theta)] = \mathrm{Ad}_{U^{-1}(\theta)}[\gamma_1(\theta),\gamma_2(\theta)] = \Phi([\gamma_1,\gamma_2])(\theta). $$
Existence of the inverse map
Define the inverse map as follows:
$$ \Psi: \gamma(\theta) \in \mathcal{L}(\mathfrak{g}) \mapsto \mathrm{Ad}_{U(\theta)} \gamma(\theta) \in \mathcal{L}(\mathfrak{g},\sigma). $$
so that fact $(2)$ ensures right inverse
$$\Phi(\Psi(\gamma))(\theta) = (\mathrm{Ad}_{U^{-1}(\theta)}\circ\mathrm{Ad}_{U(\theta)})\gamma(\theta) = \gamma(\theta)$$
and left inverse
$$\Psi(\Phi(\gamma))(\theta) = (\mathrm{Ad}_{U(\theta)}\circ \mathrm{Ad}_{U^{-1}(t)})\gamma(\theta) = \gamma(\theta)$$
Thus, we can conclude that $\mathcal{L}(\mathfrak{g}, \sigma) \cong \mathcal{L}(\mathfrak{g})$.
Can I hear your feedback? If this is reasonable, how does this extend to the affine Lie algebra?