I think I have an answer to this, but please tell me if my notation is off as well as my argument, and I will edit. The idea is to use the fact that a covering space for n-connected spaces ($n\ge 1$, i.e., connected with trivial fundamental group) introduces isomorphisms on homotopy groups.

Consider the CW-complexes $X=S^{n+1}, Y=S^{n+1+k}$ for some $k\ge 1$. Then $X$ and $Y$ both have free $\mathbb Z/2$ actions given by the antipodal map. So if we take $X\times Y$, then we can define $T_1$ to be the $\mathbb Z/2$-action on just the first factor with the identity on the second, and $T_2$ similarly acting only on the second factor.

If you quotient by $T_1$ you get $\mathbb {RP}^{n+1}\times S^{n+1+k}$ and if you quotient by $T_2$ you get $S^{n+1}\times \mathbb {RP}^{n+1+k}$. These will have all the same homotopy groups.

On the other hand, they can't be homotopy equivalent because they have different $\mathbb Z/2$ cohomology groups: the $\mathbb Z/2$ cohomology ring of $\mathbb {RP}^k$ is $\mathbb Z/2[x]/x^{k+1}$, and $S^k$ only has nontrivial $\mathbb Z/2$ cohomology at level $k$. So the cohomology of $\mathbb {RP}^{n+1}\times S^{n+1+k}$ has nontrivial elements at dimensions $1, \ldots n+1, n+1+k, \ldots n+2+k$, and the cohomology of $S^{n+1}\times \mathbb {RP}^{n+1+k}$ can have nontrivial elements at dimensions $1, \ldots n+2+k$.

EDIT: The two spaces I made are not n-connected; the homotopy groups up to n are trivial EXCEPT for the fundamental group.

The idea is to use the fact that a covering space for n-connected spaces ($n\ge 1$, i.e., connected with trivial fundamental group) introduces isomorphisms on homotopy groups.

Consider the CW-complexes $X=S^{n+1}, Y=S^{n+1+k}$ for some $k\ge 1$. Then $X$ and $Y$ both have free $\mathbb Z/2$ actions given by the antipodal map. So if we take $X\times Y$, then we can define $T_1$ to be the $\mathbb Z/2$-action on just the first factor with the identity on the second, and $T_2$ similarly acting only on the second factor.

If you quotient by $T_1$ you get $\mathbb {RP}^{n+1}\times S^{n+1+k}$ and if you quotient by $T_2$ you get $S^{n+1}\times \mathbb {RP}^{n+1+k}$. These will have all the same homotopy groups.

On the other hand, they can't be homotopy equivalent because they have different $\mathbb Z/2$ cohomology groups: the $\mathbb Z/2$ cohomology ring of $\mathbb {RP}^k$ is $\mathbb Z/2[x]/x^{k+1}$, and $S^k$ only has nontrivial $\mathbb Z/2$ cohomology at level $k$. So the cohomology of $\mathbb {RP}^{n+1}\times S^{n+1+k}$ has nontrivial elements at dimensions $1, \ldots n+1, n+1+k, \ldots n+2+k$, and the cohomology of $S^{n+1}\times \mathbb {RP}^{n+1+k}$ can have nontrivial elements at dimensions $1, \ldots n+2+k$.