Is there a way of measuring how far the group of homotopy equivalences is from the mapping class group of $X$? $[X,X]$ is the collection of homotopy equivalences from $X$ to itself, with $X$ a $CW$ complex.
To start, suppose that $X$ is a $1$-dimensional $CW$ complex. As usual, let $\mathcal{M}(X):=Homeo^+(X)/Homeo_0$ be the mapping class group.
For surfaces $\Sigma_g$ with $g \geq 1$, we know that a homotopy equivalence of $X$ with itself is homotopic to some homeomorphism, but there are spaces for which this fails, such as the wedge of two circles. In general, I want to measure how far $[X,X]$ is from $\mathcal{M}(X)$
I know that for any eilenberg-maclane space, $[X,X] \cong Out(\pi_1(X))$, so for any rose, $[X,X] \cong Out(F_n)$, whereas the mapping class group is finite. 
$Q1:$ Is $\mathcal{M}(X)$ normal in $[X,X]$, and if so, what can be said of the quotient? Is there a class of spaces that makes $\mathcal{M}(X)$ normal?
$Q2:$ Are there references for this sort of question?
$Q3:$ how about for higher dimensional $CW$ complexes?
 A: I'll turn my comments about the 1-dimensional case into a partial answer addressing only that case, and I'll add a proof that normality fails.
When $X$ is a 1-dimensional CW complex, also known to topologists as a "graph", then calculation of the mapping class group $\mathcal{M}(X)$ can be reduced to a discrete problem. Let $\mathcal{E}$ be the set of oriented edges; let $r : \mathcal{E} \to \mathcal{E}$ be the involution that reverses orientation; let $V$ be the set of vertices, and let $i : \mathcal{E} \to V$ be the map which assigns to each oriented edge its initial vertex. Then $\mathcal{M}$ is isomorphic to the group of permutations $\sigma$ of the disjoint union $\mathcal{E} \coprod V$ such that $\sigma(\mathcal{E})=\mathcal{E}$, $\sigma \circ r = r \circ \sigma$, and $\sigma \circ i = i \circ \sigma$. The proof of this boils down to the 1-dimensional version of a theorem of Alexander: every homeomorphism of $[0,1]$ that fixes $0$ and $1$ is isotopic to the identity.
To address the question of normality, the image of the natural homomorphism $\mathcal{M}(X) \to [X,X]$ is not a normal subgroup. One way to see this is to use two facts: the group $[X,X]$ acts on the set of conjugacy classes of the free group $\pi_1(X)$; and every conjugacy class is uniquely represented by a "circuit" in $X$, meaning a cyclically ordered sequence of edges without cancellation, modulo cyclic permutation. When $X$ is a finite graph, it contains only finitely many embedded circuits, corresponding to a finite set of conjugacy classes that is invariant under $\mathcal{M}(X)$. But $[X,X]$ contains an element $\Phi$ which takes an embedded circuit to a non-embedded circuit (up to homotopy), hence 
$$\Phi \mathcal{M}(X) \Phi^{-1} \ne \mathcal{M}(X)
$$
In fact if $X$ has rank $\ge 3$ one can show that $[X,X]$ contains an element $\Phi$ that has no periodic conjugacy classes. If $X$ has rank $2$, then the conjugacy class of $aba^{-1}b^{-1}$ is periodic for every element of $[X,X]$, and similarly for $ab^{-1}a^{-1}b$, and for the inverses of those two, and for all their powers. However, all you have to do is to take one embedded circuit in $X$ and find one $\Phi \in [X,X]$ which takes that embedded circuit to a nonembedded one, and that's pretty straightforward.
By the way, I like this normality question, I'm going to make it an exercise in my $\text{Out}(F_n)$ book.
