When can an automorphism of the fundamental group be written as the induced isomorphism of some self-homeomorphism? It is a common exercise to show that both automorphisms of $\pi_1(S^1)$ can be realized as induced isomorphisms of self-homeomorphisms of $S^1$. It is natural to ask if this is the case for spaces other than $S^1$.

Let $X$ be a topological space. For which $X$ and $\eta \in \operatorname{Aut}(\pi_1(X))$ is it true that $\eta = \varphi_\ast$ for some self-homeomorphism $\varphi: X \to X$?

The answer to this more general question is "all of them" where $X = \bigwedge_{1}^n S^1$ and seems to be "all of them" for $X = \sharp_1^n T^m$.
I have not yet found such a pair $(X, \eta)$ for which there is no such self-homeomorphism. 
 A: To summarize what was posted in the comments by Henrik Rüping and Misha:
There are two possible obstructions to this being true.  The first is that not every homotopy-equivalence descends to a homeomorphism.  This phenomenon is the subject of the (unsolved as of 2019) Borel conjecture (also discussed here).  The second is that not every automorphism of $\pi_1$ generates a homotopy-equivalence — there may be higher-dimensional obstructions!  
To make this concrete, consider the following examples.  
For the first obstruction, let $X=S^1\vee S^1$.  Call the generators of the fundamental groups of each copy of $S^1$ be $a$ and $b$; then by e.g. the Seifert-van Kampen theorem, $\pi_1(X)$ is the free group $\langle a,b\rangle$.  Thus $a\mapsto a, b\mapsto ab$ defines an automorphism of $\pi_1(X)$.  
But we can classify the self-homemorphisms of $X$ quite easily.  Fix a self-homoeomorphism $f$.  For each point $p\in X$, choose a small neighborhood $N_p$ and count the connected components of $N\setminus\{p\}$.  This number must be preserved by $f$, so we must have $f(o)=o$, where $o$ is the join point of $X$.  Moreover, $f$ must map the components of $N_o\setminus\{o\}$ to each other, and they must map in such a way that they can be extended to the remainder of each circle.  So the group of self-homeomorphisms is generated by the distortions of each circle and exchanging the two circles.  None of these induce $a\mapsto a, b\mapsto ab$.  
(To see that this is a homotopy-equivalence that does not descend to a homeomorphism, try to visualize how this automorphism of $\pi_1$ affects the universal cover.)  
For the second obstruction, let $f$ be a self-homeomorphism of $S^2$ with nontrivial degree $d$.  Consider $Y=S^1\times[0,1]\times S^2$, and let $X=Y/(x,0,z)\sim(x,1,f(z))$.  $X$ is a nontrivial fiber bundle of $S^2$ over the torus, $S^1\times[0,1]/(x,0)\sim(x,1)\approx S^1\times S^1$.  Now, $\pi_1(S^2)=0$, so $\pi_1(X)=\pi_1(S^1\times S^1)$.  For each circle in the torus, $\pi_1(S^1)\cong\mathbb{Z}$; let the generators be $a$ and $b$.  Then $\pi_1(X)\cong\mathbb{Z}^2$ and certainly $a\mapsto b,b\mapsto a$ is an automorphism thereof.  
The situation is reversed in degree $2$: only the 2-sphere contributes to $\pi_2$, so that $\pi_2(X)=\pi_2(S^2)\cong\mathbb{Z}$.  But now $a$ and $b$ can be distinguished: $a$ acts trivially on $\pi_2(X)$, but $b$ acts as translation by $d$.  Thus $a\mapsto b,b\mapsto a$ cannot be extended to an automorphism of $\pi_2(X)$.  
