Diffeomorphism group and isometry group of torus $T^2$ Let us consider a standard metric and differentiable structure on $T^2$.
We first consider the diffeomorphism group and isometry group of $T^2$ on itself.
1. What is the diffeomorphism group of $T^2$?
According to Wiki,
the diffeomorphism group of the torus has the homotopy-type of its linear automorphisms $S^1 \times  S^1 \times  GL(2, \mathbb{Z}).$ So is that true:
$$\mathrm{Diff}(T^2)= S^1 \times  S^1 \times  GL(2, \mathbb{Z})?$$
2. What is the isometry group of $T^2$?
From isometry group of the riemannian manifold $\mathbb{T}^2$?, it seems that there is a ambiguity whether we have
$$\mathrm{Isometry}(T^2)= D_8\rtimes_\varphi T^2?$$ where
$\varphi\colon D_8\to \mathrm{Aut}(T^2)$ with $D_8$ the order 8 dihedral group.
3. Why are $\mathrm{Diff}(T^2)$ and $\mathrm{Isometry}(T^2)$ different?
 A: Question 1 is answered in the comments. And for question 2, the formula that you copied from the other post is correct, presuming that by the "standard" metric you mean the product metric coming from the product formula $T^2 = S^1 \times S^1$ (as alluded to in the comments, there are many other interesting metrics on $T^2$, even many flat metrics, and their isometry groups can vary).
Regarding question 3, the groups $\text{Diff}(T^2)$ and $\text{Isometry}(T^2)$ are different in many ways, for many different meanings. Perhaps the simplest reason that the diffeomorphism group is never equal to the isometry group is because a diffeomorphism can "stretch" tangent vectors in a way that an isometry cannot.
In fact, on any Riemannian manifold $M$, for any point $x \in M$, and for any linear isomorphism $L : T_x M \to T_x M$, there exists a diffeomorphism $f : M \to M$ such that $f(x)=x$ and $Df_x = L : T_x M \to T_x M$ (this is an exercise in differential topology).
If you then pick a basis for $T_x M$ which is orthonormal with respect to the inner product that is specified by the given Riemannian metric on $M$, for any isometry $f : M \to M$ such that $f(x)=x$, its derivative $D_x f : T_x M \to T_x M$ is expressed in that basis as an element of the orthogonal group $\text{O}(n,\mathbb R)$, whereas the derivative at $x$ of a diffeomorphism that fixes $x$ is expressed as an element of the general linear group $\text{GL}(n,\mathbb R)$.
So all you have to do to get a diffeomorphism of $M$ that is not an isometry is to pick $L \in \text{GL}(n,\mathbb R) - \text{O}(n,\mathbb R)$.

For an independent, and somewhat different, but very interesting reason why $\text{Diff}(T^2)$ and $\text{Isometry}(T^2)$ differ: the group $\text{Diff}(T^2)$ has infinitely many path components, but no matter what Riemannian metric you pick on $T^2$ (not just the "standard" one), the image of the inclusion $\text{Isometry}(T^2) \hookrightarrow \text{Diff}(T^2)$ is contained in a union of only finitely many path components of $\text{Diff}(T^2)$. This can be expressed more clearly in algebraic topology language using the action of $\text{Diff}(T^2)$ on $H_1(T^2;\mathbb Z) \approx \mathbb Z^2$, but that's beginning to get rather far beyond the basic point of Question 3 so I'll stop now.
