Let $G$ be a finite group of diffeomorphisms of the torus $T^n$ fixing some point $p$, i.e. $p$ is fixed by every element of $G$. I have two questions.
From Edmonds' notes on Transformation groups: http://www.indiana.edu/~jfdavis/seminar/transformationgroupsb.pdf
Problem 9. [p.28] Show that if a [finite] group $G$ acts on the torus $T^n$ with a fixed point x and induces the trivial action on $\pi_1(T^n, x)$, then the action is trivial.
So the answer to question 1 is yes. Question 2 is answered by @YCor's comment and also discussed in the above notes.
This follows from a 1931 theorem of M. H. A. Newman. Theorem 1 of the paper says that for a connected manifold $M$ with a metric and an integer $m>1$, there is a constant $d$ so that any (uniformly continuous) periodic transformation of $M$ of order $m$ must move a point of $M$ at least a distance $d$.
Given a homeomorphism $f$ of a torus $T^n$ which acts trivially on $\pi_1(T^n)$, has order $m$, and fixes a point, there is a lift $\tilde{f}$ of $f$ to the universal cover $\mathbb{R}^n$ which fixes a point and has order $m$ and is uniformly continuous. Moreover, $f$ is homotopic to the identity, since $T^n$ is a $K(\mathbb{Z}^n,1)$. Then in some Euclidean metric on $T^n$, every point will be homotopic to its image under $f$ by a path of length at most $D$ (the tracks of the homotopy have bounded length). The homotopy of $f$ to the identity lifts to a homotopy of $\tilde{f}$ to the identity, so $\tilde{f}$ also has the property that it moves points at most $D$ and has order $m$. Let $d$ be the constant from Newman's theorem for the manifold $\mathbb{R}^n$ with Euclidean metric and $m$, then rescale the metric by $d/D$, we see that $\tilde{f}$ is periodic of order $m$ and moves points at most distance $d$, hence is the identity. But then $f$ is also the identity.
Hence for a finite group $G$ acting (faithfully) on a torus $T^n$, every non-trivial element $f$ of $G$ must act non-trivially on $\pi_1(T^n)$.
We can expand the answer from user83633, following hints in the notes by Edmonds, as follows.
Let $G$ be a finite group acting faithfully on an $n$-dimensional torus $T$ and fixing a point $p$. We can write $T$ as $V/L$ for some vector space $V$ and lattice $L<V$. As $T$ is homogeneous, we can assume that this identifies $p$ with the zero element. Let $\pi\colon V\to T$ be the obvious projection, which is a universal covering. Covering theory tells us that there is a unique way to let $G$ act on $V$ such that $0$ is fixed and $\pi$ is equivariant. This action preserves $\pi^{-1}\{0\}=L$, and we can identify $L$ with $H_1(T)$ equivariantly. We want to show that the action of $G$ on $L$ is faithful. If not, we can choose a subgroup $C\leq G$ of prime order $p$ that acts trivially on $L$. We will show that this leads to a contradiction.
First note that the free action map $L\times V\to V$ is $G$-equivariant, and $C$ acts trivially on $L$, so we get an induced free action of $L$ on $V^C$. The map $\pi$ gives an injection $V^C/L\to T^C$; we claim that this is a homeomorphism. As everything is compact and hausdorff, we need only check that it is surjective. Suppose that $x\in T^C$, and choose $v\in V$ with $\pi(v)=x$. If $g$ is a generator of $C$, we must then have $g.v=a+v$ for some $a\in L$. As $g$ acts trivially on $L$ this gives $g^k.v=ka+v$ for all $k$, so in particular $v=g^p.v=pa+v$ so $pa=0$ so $a=0$ so $g.v=v$ as required.
For the rest of the argument we will use various cohomology groups; these are always taken with coefficients $\mathbb{Z}/p$.
Next, recall that $H^*(BC)$ is the tensor product of a polynomial algebra on a class $x$ of degree $2$ with an exterior algebra on a class $a$ of degree $1$. For any $C$-space $X$ we have a map $EC\times_CX\to EC/C=BC$, which makes the Borel cohomology group $H_C^*(X)=H^*(EC\times_CX)$ into an algebra over $H^*(BC)$. We write $\widehat{H}^*_C(X)$ for the ring obtained by inverting $x$ in $H_C^*(X)$. If $Y$ is a $C$-subspace of $X$ then we have relative groups $\widehat{H}_C^*(X,Y)$ defined in a similar way. If $C$ acts freely on $X\setminus Y$ then these relative groups are trivial, as one can prove by cellular induction starting from the case where $X=C\times B^d$ and $Y=C\times S^{d-1}$. On the other hand, if $C$ acts trivially on $X$ then $\widehat{H}^*_C(X)=H^*(X)\otimes R^*$, where $$R^*=\widehat{H}^*_C(\text{point})=\mathbb{Z}/p[x,x^{-1}]\otimes E[a].$$ Because $C$ has prime order, it must act freely on $X\setminus X^C$, so $\widehat{H}^*_C(X,X^C)=0$, so $$ \widehat{H}^*_C(X) \simeq \widehat{H}^*_C(X^C) \simeq H^*(X^C)\otimes R^*. $$ This is essentially what is called Smith theory.
Now take $X$ to be our vector space $V$, so $V$ is contractible, and the usual spectral sequence $H^*(G;H^*(X))\Longrightarrow H^*_G(X)$ gives $H^*_G(V)=H^*(BG)$ and therefore $\widehat{H}^*_G(V)=R^*$. By comparison with Smith theory, we get $H^*(V^C)=\mathbb{Z}/p$. As $L$ acts freely on $V^C$ with $V^C/L=T^C$ we get a spectral sequence $$ H^*(L;H^*(V^C)) \Longrightarrow H^*(T^C) $$ This admits a map from the similar spectral sequence $$ H^*(L;H^*(V)) \Longrightarrow H^*(T). $$ As $H^*(V^C)=H^*(V)=\mathbb{Z}/p$, the map is an isomorphism on the initial page, and there is no room for any differentials anyway, so the map $H^*(T)\to H^*(T^C)$ is an isomorphism. In particular, the map $H^n(T)\to H^n(T^C)$ is an isomorphism.
On the other hand, if $X$ is any proper subset of $T$ then the restriction $H^n(T)\to H^n(X)$ factors through the group $H^n(T\setminus \{x\})=0$ for some $x\in T$. Thus, we must have $T^C=T$, which means that $C$ acts trivially on $T$. This contradicts our assumption that the action of $G$ is faithful.
For a simpler and more geometric proof: It suffices to show the only element which acts as the identity is the identity. Suppose $g$ is our element which acts as the identity on $H_1$ and thus on $\pi_1$. Lift $g$ to act on $\mathbb{R}^2$. Define a homotopy $g_t(p) = (1 - t) p + t g(p)$ between $g$ and the identity. The homotopy descends to $\mathbb{T}^2$. By surface theory we can upgrade this to an isotopy between $g$ and the identity on $\mathbb{T}^2$. Then, and I'm not sure how to prove this (and in some sense this seems to be the crux of the issue) one can show that maps isotopic to the identity have infinite order.
