It would'nt make sense to me to have an automorphism that does'nt preserve the trace of projections as it would send a projection to something equivalent to either a proper subprojection or greater projection but I am not able to show it for an arbitrary automorphism.
Of course for inner automorphisms it's trivial.
If this does not hold would you have a counterexample?
As mentioned by westerbaan, in any II$_1$-factor the trace is unique. This follows from the fact that the values of the trace on projections are determined by the Murray-von Neumann equivalence of projections.
Namely, let $p$ be a projection. Fix $n\in\mathbb N$. Murray and von Neumann showed that there is a Division Algorithm for projections. Namely, if you choose $n$ pairwise orthogonal equivalent projections $p_1,\ldots,p_n$ with $\sum q_j=1$, then there exists $m\leq n$, pairwise orthogonal subprojections $q_1,\ldots,q_m$ of $p$, and another subprojection $r$ of $p$ with $r\preceq q_1$, such that $$ p=\sum_{j=1}^m q_j + r. $$ Since $\tau(q_j)=1/n$, it follows that $\tau(p)=m/n+\tau(r)$, with $\tau(r)<1/n$. Thus $$ \left| \tau(p)-\frac mn\right|\leq\frac 1n. $$ As this can be done for any $n$, this determines the real number $\tau(p)$, just using the fact that $\tau$ is a trace.