Does space of Riemannian metric on Torus and on coffee mug coincide? I am very confusing about such questions : if we deform an $n$-manifold $M$ to another $M'$ (such as the torus and coffee mug) then

Does space of Riemannian metric on $M$ and on $M'$ coincide?

 A: This is a vaguely worded question whose answer depends strongly on how you choose to make it rigorous. The most rigid realization would be the following:

Q1. If $M,M'$ are diffeomorphic smooth manifolds (but possibly with different underlying sets), are their respective spaces of Riemannian metrics $\mathfrak{M}(M), \mathfrak{M}(M')$ equal?

The answer to this question is obviously no, since a metric on $M$ is a function $g: TM \times TM \to \mathbb R$ while a metric on $M'$ is a function $g':TM'\times TM' \to \mathbb R,$ so if $M,M'$ differ as sets then $g,g'$ have different domains and thus cannot be equal.
Of course, this is very rarely how mathematicians think about objects like manifolds: we don't really care about the underlying sets, since this is really just some way of formally encoding the concepts we're actually thinking about. If you're studying smooth manifolds with no additional structure, then what you really care about are diffeomorphism classes; so $M$ and $M'$ being diffeomorphic means they are essentially the same. This idea carries upwards when we construct things on top of manifolds: we should really consider $\mathfrak M(M),\mathfrak M(M')$ equal if they are isomorphic in some sense, e.g. related by a diffeomorphism. Thus a more natural formalization of your question would be something of the form:

Q2. If $M,M'$ are diffeomorphic smooth manifolds, are the spaces $\mathfrak M(M), \mathfrak M(M')$ isomorphic?

The correct sense of isomorphic could be debated, but for any sensible choice you'll find the answer is yes: if $\phi : M \to M'$ is a diffeomorphism, then the pullback map $$\phi^* : \mathfrak M(M') \to \mathfrak M(M)$$ is a bijection that will preserve any structure defined in terms of the smooth structures of $M,M'.$
This should seem obvious if you think in the right way: diffeomorphic manifolds are equal as smooth manifolds, so of course their spaces of metric will be equal. 
