Preserving mutual information after compressing states Let $X$ and $Y$ be stochastic variables on respectively $n$ and $m$ points with $m>n$ and a joint probability distribution $p(x,y)$.
The mutual information is
$$
I(X ;Y) = H(X) + H(Y) - H(X,Y)
$$
where $H(X)$ denotes the Shannon entropy of the marginal of $p$ over $X$ and $H(X,Y)$ is the Shannon entropy of the joint distribution $p$.
Is it possible to compress $Y$ to the size of $X$ whilst preserving mutual information? That is, does there exist a stochastic matrix $T: \mathbb{R}^m \rightarrow \mathbb{R}^n$ which sends $p$ to $(I_n\otimes T)p$, such that
$$
I(X;Y) = I(X;Y^\prime)
$$
Intuitively this makes sense as the maximal amount of information that they can share should depend on the smallest dimension of the two. I however couldn't find any result like this.
 A: Ahlswede and Gacs showed the following 'strong data processing inequality' in the mid-70s.

Suppose the channel $P_{U|V}$ is such that $0$ error communication is not possible over it - i.e., for every $v \neq v', \exists u: P(u|v)P(u|v')> 0$. Then there exists an $\eta < 1$ such that for any Markov chain $W-V-U,$ $$I(W;U) \le \eta I(V;U).$$

In our case, let $W$ be the quantised version of $Y$, $V=Y$ and $U= X$. If the channel $P_{X|Y}$ cannot have $0$ error communication, then it follows that we must lose some information in the quantisation. So, basically anything can serve as a counterexample (although it's non-trivial to see how). 
Strictly speaking the inequality above is quite a big hammer - for instance, it shows that varying the distribution of $Y$ while keeping the channel $P_{X|Y}$ fixed cannot help. 
See this recent survey due to Polyanskiy and Wu for an account of strong data processing inequalities. The inequality I cite is equation (21) there. 

An alternate: 
In a comment you had asked how few levels $Y$ needed to be quantised to in order to attain good compression of the information about $X$. This paper studies a slightly different problem, which you may be interested in. In short: consider the set of random variables $X,Y$ with joint distribution such that $I(X;Y) \ge \beta$. They study the minimax question of how much information a $M<|\mathcal{Y}|$ level quantiser can retain about $X$ in the worst case as one varies $P_{XY}$ subject to the prior constraint. In fact, with their converse results, you can find other counterexamples for your problem - for instance, they show that for binary $X$, there exists some joint distribution $P_{XY}$ such that any binary quantisation $Y_2$ of $Y$ must have $I(X;Y_2) \le 3\beta/\max( \log(1/\beta), 1). $ This means that there should exist distributions with mutual information $< 1/8$ for which you cannot retain sufficent mutual information in any $2$-level quantisation (possibly you can figure something out explicitly using the constructions in their upper bound). 
A: Here's my attempt at answering this interesting question.
The random variables  $X,Y,T(Y)$ form a Markov chain $X\rightarrow Y \rightarrow T(Y)$. By the data processing inequality, it always holds
$$
I(X;T(Y))\leq I(X;Y)
$$
with equality if and only if it also holds $X\rightarrow T(Y) \rightarrow Y$. 
The latter condition is what defines the so called sufficient statistic in estimation theory [Cover&Thomas, Ch. 2]. Therefore, your question may be equivalently posed as follows: Is it always possible to find a sufficient statistic $T(Y)$ of dimension smaller than the dimension of the "parameter" $X$?
It turns out that this is not always possible. Consider the following example (taken from these slides). $X\in \mathbb{R}$ is an one-dimensional random variable (of some arbitrary distribution) and $Y\in \mathbb{R}^n$ is an $m$-dimensional random variable whose elements are i.i.d. uniformly distributed over the interval $[X,X+1]$. It can be shown that the so-called minimal sufficient statistic in this case is the two-dimensional vector $(\min\{Y_i\},\max\{Y_i\})$. Therefore, although "compression" of the observation is possible, the dimension of the minimal sufficient statistic is greater than that of $X$. Since the transform $T$ is non-linear in this case, it follows that restricting our attention to linear transforms can only result in an increase of the sufficient statistics dimension.
